Mixing
Show $a_n$ is unbounded if $a_n= a_{n-1} left(1+ frac{1}{sqrt n}right)$ to determine that $a_n$ diverges.
$begingroup$
I wish to determine the limit of $a_n$, where we recursively define:
$$a_n= a_{n-1} left(1+ frac{1}{sqrt n}right)$$ where $a_0=1$
I already noticed it is increasing, because
$$a_n - a_{n-1}= frac{1}{sqrt{n}}a_{n-1}>0 $$
Since all terms are strictly positive. (alternatively, we could form a better argument via induction).
edit:
Base case: $a_1 = 2 > 1 = a_0$.
Hypothesis: $a_k > a_{k-1}$
Step: $a _ {k+1}= a_{k} left(1+ frac{1}{sqrt n}right)>a_{k-1} left(1+ frac{1}{sqrt n}right)=a_k $
Therefore $a_n >a_{n-1}$ for all $n$ and the sequence is increasing.
I plotted some terms and this seems to grow exponentially, how would I prove it diverges? I think I need to show it is unbounded.I tried the ratio test for sequences but this is inconclusive as the ratio tends to $1$.
real-analysis sequences-and-series divergent-series
edited Mar 10 at 20:10
Robert Howard
2,2383935
asked Jan 27 at 13:00
Wesley StrikWesley Strik
2,209424
$endgroup$
|
show 5 more comments
$begingroup$
I wish to determine the limit of $a_n$, where we recursively define:
$$a_n= a_{n-1} left(1+ frac{1}{sqrt n}right)$$ where $a_0=1$
I already noticed it is increasing, because
$$a_n - a_{n-1}= frac{1}{sqrt{n}}a_{n-1}>0 $$
Since all terms are strictly positive. (alternatively, we could form a better argument via induction).
edit:
Base case: $a_1 = 2 > 1 = a_0$.
Hypothesis: $a_k > a_{k-1}$
Step: $a _ {k+1}= a_{k} left(1+ frac{1}{sqrt n}right)>a_{k-1} left(1+ frac{1}{sqrt n}right)=a_k $
Therefore $a_n >a_{n-1}$ for all $n$ and the sequence is increasing.
I plotted some terms and this seems to grow exponentially, how would I prove it diverges? I think I need to show it is unbounded.I tried the ratio test for sequences but this is inconclusive as the ratio tends to $1$.
real-analysis sequences-and-series divergent-series
edited Mar 10 at 20:10
Robert Howard
2,2383935
asked Jan 27 at 13:00
Wesley StrikWesley Strik
2,209424
$endgroup$
2
$begingroup$
Think about $log a_n$.
$endgroup$
– Lord Shark the Unknown
Jan 27 at 13:03
$begingroup$
It doesn't look very exponential from this angle...
$endgroup$
– Rhys
Jan 27 at 13:05
$begingroup$
wolframalpha.com/input/…
$endgroup$
– Wesley Strik
Jan 27 at 13:05
1
$begingroup$
You’re asking whether the infinite product $prod(1+1/sqrt n)$ converges. This question should be closely relatable to the convergence of $sum1/sqrt n$.
$endgroup$
– Lubin
Jan 27 at 13:17
1
$begingroup$
Hint $$a_n = a_{n-1}left(1+frac{1}{sqrt{n}}right)=a_{n-2}left(1+frac{1}{sqrt{n-1}}right)left(1+frac{1}{sqrt{n}}right)=\ a_0prodlimits_{k=1}^nleft(1+frac{1}{sqrt{k}}right)geq left(1+frac{1}{sqrt{n}}right)^n= left(left(1+frac{1}{sqrt{n}}right)^{sqrt{n}}right)^{sqrt{n}} approx e^{sqrt{n}}$$
$endgroup$
– rtybase
Jan 27 at 13:31
|
show 5 more comments
$begingroup$
I wish to determine the limit of $a_n$, where we recursively define:
$$a_n= a_{n-1} left(1+ frac{1}{sqrt n}right)$$ where $a_0=1$
I already noticed it is increasing, because
$$a_n - a_{n-1}= frac{1}{sqrt{n}}a_{n-1}>0 $$
Since all terms are strictly positive. (alternatively, we could form a better argument via induction).
edit:
Base case: $a_1 = 2 > 1 = a_0$.
Hypothesis: $a_k > a_{k-1}$
Step: $a _ {k+1}= a_{k} left(1+ frac{1}{sqrt n}right)>a_{k-1} left(1+ frac{1}{sqrt n}right)=a_k $
Therefore $a_n >a_{n-1}$ for all $n$ and the sequence is increasing.
I plotted some terms and this seems to grow exponentially, how would I prove it diverges? I think I need to show it is unbounded.I tried the ratio test for sequences but this is inconclusive as the ratio tends to $1$.
real-analysis sequences-and-series divergent-series
edited Mar 10 at 20:10
Robert Howard
2,2383935
asked Jan 27 at 13:00
Wesley StrikWesley Strik
2,209424
$endgroup$
I wish to determine the limit of $a_n$, where we recursively define:
$$a_n= a_{n-1} left(1+ frac{1}{sqrt n}right)$$ where $a_0=1$
I already noticed it is increasing, because
$$a_n - a_{n-1}= frac{1}{sqrt{n}}a_{n-1}>0 $$
Since all terms are strictly positive. (alternatively, we could form a better argument via induction).
edit:
Base case: $a_1 = 2 > 1 = a_0$.
Hypothesis: $a_k > a_{k-1}$
Step: $a _ {k+1}= a_{k} left(1+ frac{1}{sqrt n}right)>a_{k-1} left(1+ frac{1}{sqrt n}right)=a_k $
Therefore $a_n >a_{n-1}$ for all $n$ and the sequence is increasing.
I plotted some terms and this seems to grow exponentially, how would I prove it diverges? I think I need to show it is unbounded.I tried the ratio test for sequences but this is inconclusive as the ratio tends to $1$.
real-analysis sequences-and-series divergent-series
real-analysis sequences-and-series divergent-series
edited Mar 10 at 20:10
Robert Howard
2,2383935
asked Jan 27 at 13:00
Wesley StrikWesley Strik
2,209424
edited Mar 10 at 20:10
Robert Howard
2,2383935
asked Jan 27 at 13:00
Wesley StrikWesley Strik
2,209424
edited Mar 10 at 20:10
Robert Howard
2,2383935
edited Mar 10 at 20:10
Robert Howard
2,2383935
edited Mar 10 at 20:10
Robert Howard
2,2383935
2,2383935
asked Jan 27 at 13:00
Wesley StrikWesley Strik
2,209424
asked Jan 27 at 13:00
Wesley StrikWesley Strik
2,209424
asked Jan 27 at 13:00
Wesley StrikWesley Strik
2,209424
2,209424
2
$begingroup$
Think about $log a_n$.
$endgroup$
– Lord Shark the Unknown
Jan 27 at 13:03
$begingroup$
It doesn't look very exponential from this angle...
$endgroup$
– Rhys
Jan 27 at 13:05
$begingroup$
wolframalpha.com/input/…
$endgroup$
– Wesley Strik
Jan 27 at 13:05
1
$begingroup$
You’re asking whether the infinite product $prod(1+1/sqrt n)$ converges. This question should be closely relatable to the convergence of $sum1/sqrt n$.
$endgroup$
– Lubin
Jan 27 at 13:17
1
$begingroup$
Hint $$a_n = a_{n-1}left(1+frac{1}{sqrt{n}}right)=a_{n-2}left(1+frac{1}{sqrt{n-1}}right)left(1+frac{1}{sqrt{n}}right)=\ a_0prodlimits_{k=1}^nleft(1+frac{1}{sqrt{k}}right)geq left(1+frac{1}{sqrt{n}}right)^n= left(left(1+frac{1}{sqrt{n}}right)^{sqrt{n}}right)^{sqrt{n}} approx e^{sqrt{n}}$$
$endgroup$
– rtybase
Jan 27 at 13:31
|
show 5 more comments
2
$begingroup$
Think about $log a_n$.
$endgroup$
– Lord Shark the Unknown
Jan 27 at 13:03
$begingroup$
It doesn't look very exponential from this angle...
$endgroup$
– Rhys
Jan 27 at 13:05
$begingroup$
wolframalpha.com/input/…
$endgroup$
– Wesley Strik
Jan 27 at 13:05
1
$begingroup$
You’re asking whether the infinite product $prod(1+1/sqrt n)$ converges. This question should be closely relatable to the convergence of $sum1/sqrt n$.
$endgroup$
– Lubin
Jan 27 at 13:17
1
$begingroup$
Hint $$a_n = a_{n-1}left(1+frac{1}{sqrt{n}}right)=a_{n-2}left(1+frac{1}{sqrt{n-1}}right)left(1+frac{1}{sqrt{n}}right)=\ a_0prodlimits_{k=1}^nleft(1+frac{1}{sqrt{k}}right)geq left(1+frac{1}{sqrt{n}}right)^n= left(left(1+frac{1}{sqrt{n}}right)^{sqrt{n}}right)^{sqrt{n}} approx e^{sqrt{n}}$$
$endgroup$
– rtybase
Jan 27 at 13:31
2
2
$begingroup$
Think about $log a_n$.
$endgroup$
– Lord Shark the Unknown
Jan 27 at 13:03
$begingroup$
Think about $log a_n$.
$endgroup$
– Lord Shark the Unknown
Jan 27 at 13:03
$begingroup$
It doesn't look very exponential from this angle...
$endgroup$
– Rhys
Jan 27 at 13:05
$begingroup$
It doesn't look very exponential from this angle...
$endgroup$
– Rhys
Jan 27 at 13:05
$begingroup$
wolframalpha.com/input/…
$endgroup$
– Wesley Strik
Jan 27 at 13:05
$begingroup$
wolframalpha.com/input/…
$endgroup$
– Wesley Strik
Jan 27 at 13:05
1
1
$begingroup$
You’re asking whether the infinite product $prod(1+1/sqrt n)$ converges. This question should be closely relatable to the convergence of $sum1/sqrt n$.
$endgroup$
– Lubin
Jan 27 at 13:17
$begingroup$
You’re asking whether the infinite product $prod(1+1/sqrt n)$ converges. This question should be closely relatable to the convergence of $sum1/sqrt n$.
$endgroup$
– Lubin
Jan 27 at 13:17
1
1
$begingroup$
Hint $$a_n = a_{n-1}left(1+frac{1}{sqrt{n}}right)=a_{n-2}left(1+frac{1}{sqrt{n-1}}right)left(1+frac{1}{sqrt{n}}right)=\ a_0prodlimits_{k=1}^nleft(1+frac{1}{sqrt{k}}right)geq left(1+frac{1}{sqrt{n}}right)^n= left(left(1+frac{1}{sqrt{n}}right)^{sqrt{n}}right)^{sqrt{n}} approx e^{sqrt{n}}$$
$endgroup$
– rtybase
Jan 27 at 13:31
$begingroup$
Hint $$a_n = a_{n-1}left(1+frac{1}{sqrt{n}}right)=a_{n-2}left(1+frac{1}{sqrt{n-1}}right)left(1+frac{1}{sqrt{n}}right)=\ a_0prodlimits_{k=1}^nleft(1+frac{1}{sqrt{k}}right)geq left(1+frac{1}{sqrt{n}}right)^n= left(left(1+frac{1}{sqrt{n}}right)^{sqrt{n}}right)^{sqrt{n}} approx e^{sqrt{n}}$$
$endgroup$
– rtybase
Jan 27 at 13:31
|
show 5 more comments
4 Answers
4
active
oldest
votes
$begingroup$
Further to my comment, starting with $$a_n = a_{n-1}left(1+frac{1}{sqrt{n}}right)=a_{n-2}left(1+frac{1}{sqrt{n-1}}right)left(1+frac{1}{sqrt{n}}right)=\
a_0prodlimits_{k=1}^nleft(1+frac{1}{sqrt{k}}right)geq left(1+frac{1}{sqrt{n}}right)^n=
left(left(1+frac{1}{sqrt{n}}right)^{sqrt{n}}right)^{sqrt{n}} geq ...$$
and applying Bernoulli's inequality
$$...geq left(1+frac{sqrt{n}}{sqrt{n}}right)^{sqrt{n}}=2^{sqrt{n}}$$
As a result
$$a_n geq 2^{sqrt{n}}$$
and the result follows, i.e. the sequence is unbounded and diverges.
answered Jan 27 at 13:39
rtybasertybase
11.5k31534
$endgroup$
$begingroup$
So the trick is to estimate every term by the smallest term and show that we still get divergence either by Bernoulli's inequality or by comparison with $e^{sqrt{n}}$
$endgroup$
– Wesley Strik
Jan 27 at 13:48
1
$begingroup$
Yes, it's a form of a squeeze theorem
$endgroup$
– rtybase
Jan 27 at 13:50
1
$begingroup$
$e$xcellent answer. It feels more like a limit comparison result.
$endgroup$
– Wesley Strik
Jan 27 at 13:50
add a comment |
$begingroup$
Use $sqrt n + 1 geq sqrt{n+1}, ninmathbb N:$
$$a_n = a_{n-1}(1+frac 1{sqrt n})implies a_n = prod_{k=1}^n(1+frac 1{sqrt k})= prod_{k=1}^n frac{sqrt k + 1}{sqrt k} geq prod_{k=1}^nfrac{sqrt{k+1}}{sqrt k} = sqrt{n+1}.$$
answered Jan 27 at 14:17
EnnarEnnar
14.8k32445
$endgroup$
$begingroup$
That is a satisfying approach.
$endgroup$
– Wesley Strik
Jan 27 at 16:09
add a comment |
$begingroup$
$a_n=prod_{i=1}^nleft(1+frac1{sqrt{i}}right).$ Now notice that $f(x)=1+x-e^{x/4}ge 0$ for $xleq 1.$ Hence $a_ngeq e^{frac14sum_{i=1}^nfrac1{sqrt{i}}}rightarrow infty.$
answered Jan 27 at 13:44
John_WickJohn_Wick
1,616111
$endgroup$
$begingroup$
Clever, I would need some more practice to come up with something like this.
$endgroup$
– Wesley Strik
Jan 27 at 13:52
add a comment |
$begingroup$
Another approach:
$$a_n=a_{n-1}+frac{1}{sqrt{n}}a_{n-1}geq a_{n-1}+frac{1}{sqrt{n}},$$
from which we quickly get
$$a_ngeqsum_{k=1}^nfrac{1}{sqrt{k}}geqsum_{k=1}^nfrac{1}{k}$$
which is divergent, since the last sum gives the harmonic series.
answered Jan 27 at 14:05
WojowuWojowu
19.2k23174
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Further to my comment, starting with $$a_n = a_{n-1}left(1+frac{1}{sqrt{n}}right)=a_{n-2}left(1+frac{1}{sqrt{n-1}}right)left(1+frac{1}{sqrt{n}}right)=\
a_0prodlimits_{k=1}^nleft(1+frac{1}{sqrt{k}}right)geq left(1+frac{1}{sqrt{n}}right)^n=
left(left(1+frac{1}{sqrt{n}}right)^{sqrt{n}}right)^{sqrt{n}} geq ...$$
and applying Bernoulli's inequality
$$...geq left(1+frac{sqrt{n}}{sqrt{n}}right)^{sqrt{n}}=2^{sqrt{n}}$$
As a result
$$a_n geq 2^{sqrt{n}}$$
and the result follows, i.e. the sequence is unbounded and diverges.
answered Jan 27 at 13:39
rtybasertybase
11.5k31534
$endgroup$
$begingroup$
So the trick is to estimate every term by the smallest term and show that we still get divergence either by Bernoulli's inequality or by comparison with $e^{sqrt{n}}$
$endgroup$
– Wesley Strik
Jan 27 at 13:48
1
$begingroup$
Yes, it's a form of a squeeze theorem
$endgroup$
– rtybase
Jan 27 at 13:50
1
$begingroup$
$e$xcellent answer. It feels more like a limit comparison result.
$endgroup$
– Wesley Strik
Jan 27 at 13:50
add a comment |
$begingroup$
Further to my comment, starting with $$a_n = a_{n-1}left(1+frac{1}{sqrt{n}}right)=a_{n-2}left(1+frac{1}{sqrt{n-1}}right)left(1+frac{1}{sqrt{n}}right)=\
a_0prodlimits_{k=1}^nleft(1+frac{1}{sqrt{k}}right)geq left(1+frac{1}{sqrt{n}}right)^n=
left(left(1+frac{1}{sqrt{n}}right)^{sqrt{n}}right)^{sqrt{n}} geq ...$$
and applying Bernoulli's inequality
$$...geq left(1+frac{sqrt{n}}{sqrt{n}}right)^{sqrt{n}}=2^{sqrt{n}}$$
As a result
$$a_n geq 2^{sqrt{n}}$$
and the result follows, i.e. the sequence is unbounded and diverges.
answered Jan 27 at 13:39
rtybasertybase
11.5k31534
$endgroup$
$begingroup$
So the trick is to estimate every term by the smallest term and show that we still get divergence either by Bernoulli's inequality or by comparison with $e^{sqrt{n}}$
$endgroup$
– Wesley Strik
Jan 27 at 13:48
1
$begingroup$
Yes, it's a form of a squeeze theorem
$endgroup$
– rtybase
Jan 27 at 13:50
1
$begingroup$
$e$xcellent answer. It feels more like a limit comparison result.
$endgroup$
– Wesley Strik
Jan 27 at 13:50
add a comment |
$begingroup$
Further to my comment, starting with $$a_n = a_{n-1}left(1+frac{1}{sqrt{n}}right)=a_{n-2}left(1+frac{1}{sqrt{n-1}}right)left(1+frac{1}{sqrt{n}}right)=\
a_0prodlimits_{k=1}^nleft(1+frac{1}{sqrt{k}}right)geq left(1+frac{1}{sqrt{n}}right)^n=
left(left(1+frac{1}{sqrt{n}}right)^{sqrt{n}}right)^{sqrt{n}} geq ...$$
and applying Bernoulli's inequality
$$...geq left(1+frac{sqrt{n}}{sqrt{n}}right)^{sqrt{n}}=2^{sqrt{n}}$$
As a result
$$a_n geq 2^{sqrt{n}}$$
and the result follows, i.e. the sequence is unbounded and diverges.
answered Jan 27 at 13:39
rtybasertybase
11.5k31534
$endgroup$
Further to my comment, starting with $$a_n = a_{n-1}left(1+frac{1}{sqrt{n}}right)=a_{n-2}left(1+frac{1}{sqrt{n-1}}right)left(1+frac{1}{sqrt{n}}right)=\
a_0prodlimits_{k=1}^nleft(1+frac{1}{sqrt{k}}right)geq left(1+frac{1}{sqrt{n}}right)^n=
left(left(1+frac{1}{sqrt{n}}right)^{sqrt{n}}right)^{sqrt{n}} geq ...$$
and applying Bernoulli's inequality
$$...geq left(1+frac{sqrt{n}}{sqrt{n}}right)^{sqrt{n}}=2^{sqrt{n}}$$
As a result
$$a_n geq 2^{sqrt{n}}$$
and the result follows, i.e. the sequence is unbounded and diverges.
answered Jan 27 at 13:39
rtybasertybase
11.5k31534
edited Jan 27 at 13:47
answered Jan 27 at 13:39
rtybasertybase
11.5k31534
answered Jan 27 at 13:39
rtybasertybase
11.5k31534
answered Jan 27 at 13:39
rtybasertybase
11.5k31534
11.5k31534
$begingroup$
So the trick is to estimate every term by the smallest term and show that we still get divergence either by Bernoulli's inequality or by comparison with $e^{sqrt{n}}$
$endgroup$
– Wesley Strik
Jan 27 at 13:48
1
$begingroup$
Yes, it's a form of a squeeze theorem
$endgroup$
– rtybase
Jan 27 at 13:50
1
$begingroup$
$e$xcellent answer. It feels more like a limit comparison result.
$endgroup$
– Wesley Strik
Jan 27 at 13:50
add a comment |
$begingroup$
So the trick is to estimate every term by the smallest term and show that we still get divergence either by Bernoulli's inequality or by comparison with $e^{sqrt{n}}$
$endgroup$
– Wesley Strik
Jan 27 at 13:48
1
$begingroup$
Yes, it's a form of a squeeze theorem
$endgroup$
– rtybase
Jan 27 at 13:50
1
$begingroup$
$e$xcellent answer. It feels more like a limit comparison result.
$endgroup$
– Wesley Strik
Jan 27 at 13:50
$begingroup$
So the trick is to estimate every term by the smallest term and show that we still get divergence either by Bernoulli's inequality or by comparison with $e^{sqrt{n}}$
$endgroup$
– Wesley Strik
Jan 27 at 13:48
$begingroup$
So the trick is to estimate every term by the smallest term and show that we still get divergence either by Bernoulli's inequality or by comparison with $e^{sqrt{n}}$
$endgroup$
– Wesley Strik
Jan 27 at 13:48
1
1
$begingroup$
Yes, it's a form of a squeeze theorem
$endgroup$
– rtybase
Jan 27 at 13:50
$begingroup$
Yes, it's a form of a squeeze theorem
$endgroup$
– rtybase
Jan 27 at 13:50
1
1
$begingroup$
$e$xcellent answer. It feels more like a limit comparison result.
$endgroup$
– Wesley Strik
Jan 27 at 13:50
$begingroup$
$e$xcellent answer. It feels more like a limit comparison result.
$endgroup$
– Wesley Strik
Jan 27 at 13:50
add a comment |
$begingroup$
Use $sqrt n + 1 geq sqrt{n+1}, ninmathbb N:$
$$a_n = a_{n-1}(1+frac 1{sqrt n})implies a_n = prod_{k=1}^n(1+frac 1{sqrt k})= prod_{k=1}^n frac{sqrt k + 1}{sqrt k} geq prod_{k=1}^nfrac{sqrt{k+1}}{sqrt k} = sqrt{n+1}.$$
answered Jan 27 at 14:17
EnnarEnnar
14.8k32445
$endgroup$
$begingroup$
That is a satisfying approach.
$endgroup$
– Wesley Strik
Jan 27 at 16:09
add a comment |
$begingroup$
Use $sqrt n + 1 geq sqrt{n+1}, ninmathbb N:$
$$a_n = a_{n-1}(1+frac 1{sqrt n})implies a_n = prod_{k=1}^n(1+frac 1{sqrt k})= prod_{k=1}^n frac{sqrt k + 1}{sqrt k} geq prod_{k=1}^nfrac{sqrt{k+1}}{sqrt k} = sqrt{n+1}.$$
answered Jan 27 at 14:17
EnnarEnnar
14.8k32445
$endgroup$
$begingroup$
That is a satisfying approach.
$endgroup$
– Wesley Strik
Jan 27 at 16:09
add a comment |
$begingroup$
Use $sqrt n + 1 geq sqrt{n+1}, ninmathbb N:$
$$a_n = a_{n-1}(1+frac 1{sqrt n})implies a_n = prod_{k=1}^n(1+frac 1{sqrt k})= prod_{k=1}^n frac{sqrt k + 1}{sqrt k} geq prod_{k=1}^nfrac{sqrt{k+1}}{sqrt k} = sqrt{n+1}.$$
answered Jan 27 at 14:17
EnnarEnnar
14.8k32445
$endgroup$
Use $sqrt n + 1 geq sqrt{n+1}, ninmathbb N:$
$$a_n = a_{n-1}(1+frac 1{sqrt n})implies a_n = prod_{k=1}^n(1+frac 1{sqrt k})= prod_{k=1}^n frac{sqrt k + 1}{sqrt k} geq prod_{k=1}^nfrac{sqrt{k+1}}{sqrt k} = sqrt{n+1}.$$
answered Jan 27 at 14:17
EnnarEnnar
14.8k32445
answered Jan 27 at 14:17
EnnarEnnar
14.8k32445
answered Jan 27 at 14:17
EnnarEnnar
14.8k32445
answered Jan 27 at 14:17
EnnarEnnar
14.8k32445
14.8k32445
$begingroup$
That is a satisfying approach.
$endgroup$
– Wesley Strik
Jan 27 at 16:09
add a comment |
$begingroup$
That is a satisfying approach.
$endgroup$
– Wesley Strik
Jan 27 at 16:09
$begingroup$
That is a satisfying approach.
$endgroup$
– Wesley Strik
Jan 27 at 16:09
$begingroup$
That is a satisfying approach.
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– Wesley Strik
Jan 27 at 16:09
add a comment |
$begingroup$
$a_n=prod_{i=1}^nleft(1+frac1{sqrt{i}}right).$ Now notice that $f(x)=1+x-e^{x/4}ge 0$ for $xleq 1.$ Hence $a_ngeq e^{frac14sum_{i=1}^nfrac1{sqrt{i}}}rightarrow infty.$
answered Jan 27 at 13:44
John_WickJohn_Wick
1,616111
$endgroup$
$begingroup$
Clever, I would need some more practice to come up with something like this.
$endgroup$
– Wesley Strik
Jan 27 at 13:52
add a comment |
$begingroup$
$a_n=prod_{i=1}^nleft(1+frac1{sqrt{i}}right).$ Now notice that $f(x)=1+x-e^{x/4}ge 0$ for $xleq 1.$ Hence $a_ngeq e^{frac14sum_{i=1}^nfrac1{sqrt{i}}}rightarrow infty.$
answered Jan 27 at 13:44
John_WickJohn_Wick
1,616111
$endgroup$
$begingroup$
Clever, I would need some more practice to come up with something like this.
$endgroup$
– Wesley Strik
Jan 27 at 13:52
add a comment |
$begingroup$
$a_n=prod_{i=1}^nleft(1+frac1{sqrt{i}}right).$ Now notice that $f(x)=1+x-e^{x/4}ge 0$ for $xleq 1.$ Hence $a_ngeq e^{frac14sum_{i=1}^nfrac1{sqrt{i}}}rightarrow infty.$
answered Jan 27 at 13:44
John_WickJohn_Wick
1,616111
$endgroup$
$a_n=prod_{i=1}^nleft(1+frac1{sqrt{i}}right).$ Now notice that $f(x)=1+x-e^{x/4}ge 0$ for $xleq 1.$ Hence $a_ngeq e^{frac14sum_{i=1}^nfrac1{sqrt{i}}}rightarrow infty.$
answered Jan 27 at 13:44
John_WickJohn_Wick
1,616111
answered Jan 27 at 13:44
John_WickJohn_Wick
1,616111
answered Jan 27 at 13:44
John_WickJohn_Wick
1,616111
answered Jan 27 at 13:44
John_WickJohn_Wick
1,616111
1,616111
$begingroup$
Clever, I would need some more practice to come up with something like this.
$endgroup$
– Wesley Strik
Jan 27 at 13:52
add a comment |
$begingroup$
Clever, I would need some more practice to come up with something like this.
$endgroup$
– Wesley Strik
Jan 27 at 13:52
$begingroup$
Clever, I would need some more practice to come up with something like this.
$endgroup$
– Wesley Strik
Jan 27 at 13:52
$begingroup$
Clever, I would need some more practice to come up with something like this.
$endgroup$
– Wesley Strik
Jan 27 at 13:52
add a comment |
$begingroup$
Another approach:
$$a_n=a_{n-1}+frac{1}{sqrt{n}}a_{n-1}geq a_{n-1}+frac{1}{sqrt{n}},$$
from which we quickly get
$$a_ngeqsum_{k=1}^nfrac{1}{sqrt{k}}geqsum_{k=1}^nfrac{1}{k}$$
which is divergent, since the last sum gives the harmonic series.
answered Jan 27 at 14:05
WojowuWojowu
19.2k23174
$endgroup$
add a comment |
$begingroup$
Another approach:
$$a_n=a_{n-1}+frac{1}{sqrt{n}}a_{n-1}geq a_{n-1}+frac{1}{sqrt{n}},$$
from which we quickly get
$$a_ngeqsum_{k=1}^nfrac{1}{sqrt{k}}geqsum_{k=1}^nfrac{1}{k}$$
which is divergent, since the last sum gives the harmonic series.
answered Jan 27 at 14:05
WojowuWojowu
19.2k23174
$endgroup$
add a comment |
$begingroup$
Another approach:
$$a_n=a_{n-1}+frac{1}{sqrt{n}}a_{n-1}geq a_{n-1}+frac{1}{sqrt{n}},$$
from which we quickly get
$$a_ngeqsum_{k=1}^nfrac{1}{sqrt{k}}geqsum_{k=1}^nfrac{1}{k}$$
which is divergent, since the last sum gives the harmonic series.
answered Jan 27 at 14:05
WojowuWojowu
19.2k23174
$endgroup$
Another approach:
$$a_n=a_{n-1}+frac{1}{sqrt{n}}a_{n-1}geq a_{n-1}+frac{1}{sqrt{n}},$$
from which we quickly get
$$a_ngeqsum_{k=1}^nfrac{1}{sqrt{k}}geqsum_{k=1}^nfrac{1}{k}$$
which is divergent, since the last sum gives the harmonic series.
answered Jan 27 at 14:05
WojowuWojowu
19.2k23174
answered Jan 27 at 14:05
WojowuWojowu
19.2k23174
answered Jan 27 at 14:05
WojowuWojowu
19.2k23174
answered Jan 27 at 14:05
WojowuWojowu
19.2k23174
19.2k23174
add a comment |
add a comment |
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2
$begingroup$
Think about $log a_n$.
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– Lord Shark the Unknown
Jan 27 at 13:03
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It doesn't look very exponential from this angle...
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– Rhys
Jan 27 at 13:05
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wolframalpha.com/input/…
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– Wesley Strik
Jan 27 at 13:05
1
$begingroup$
You’re asking whether the infinite product $prod(1+1/sqrt n)$ converges. This question should be closely relatable to the convergence of $sum1/sqrt n$.
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– Lubin
Jan 27 at 13:17
1
$begingroup$
Hint $$a_n = a_{n-1}left(1+frac{1}{sqrt{n}}right)=a_{n-2}left(1+frac{1}{sqrt{n-1}}right)left(1+frac{1}{sqrt{n}}right)=\ a_0prodlimits_{k=1}^nleft(1+frac{1}{sqrt{k}}right)geq left(1+frac{1}{sqrt{n}}right)^n= left(left(1+frac{1}{sqrt{n}}right)^{sqrt{n}}right)^{sqrt{n}} approx e^{sqrt{n}}$$
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– rtybase
Jan 27 at 13:31