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Determine the convergence of $sum_{n=1}^infty frac{4^n+n}{n!}$ [duplicate]
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This question already has an answer here:
Use any method, determine whether the series $sum_{k=1}^{infty} frac{7^{k} + k}{k! + 1}$ converges?
4 answers
$sum_{n=1}^infty frac{4^n+n}{n!}$
Using ratio test :
$lim_{n to infty} frac{a_{n+1}}{a_n}\
lim_{n to infty} frac{(4^{n+1}+n+1)n!}{(n+1)!(4^n+n)}\
lim_{n to infty} frac{(4^{n+1}+n+1)}{(n+1)(4^n+n)}$
I think it's converges, but I didn't know how to solve the limit.
calculus sequences-and-series
asked Feb 3 at 9:48
airlanggaairlangga
775
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marked as duplicate by Nosrati, Community♦ Feb 3 at 10:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Use any method, determine whether the series $sum_{k=1}^{infty} frac{7^{k} + k}{k! + 1}$ converges?
4 answers
$sum_{n=1}^infty frac{4^n+n}{n!}$
Using ratio test :
$lim_{n to infty} frac{a_{n+1}}{a_n}\
lim_{n to infty} frac{(4^{n+1}+n+1)n!}{(n+1)!(4^n+n)}\
lim_{n to infty} frac{(4^{n+1}+n+1)}{(n+1)(4^n+n)}$
I think it's converges, but I didn't know how to solve the limit.
calculus sequences-and-series
asked Feb 3 at 9:48
airlanggaairlangga
775
$endgroup$
marked as duplicate by Nosrati, Community♦ Feb 3 at 10:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Split ot into two series.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 3 at 9:50
add a comment |
$begingroup$
This question already has an answer here:
Use any method, determine whether the series $sum_{k=1}^{infty} frac{7^{k} + k}{k! + 1}$ converges?
4 answers
$sum_{n=1}^infty frac{4^n+n}{n!}$
Using ratio test :
$lim_{n to infty} frac{a_{n+1}}{a_n}\
lim_{n to infty} frac{(4^{n+1}+n+1)n!}{(n+1)!(4^n+n)}\
lim_{n to infty} frac{(4^{n+1}+n+1)}{(n+1)(4^n+n)}$
I think it's converges, but I didn't know how to solve the limit.
calculus sequences-and-series
asked Feb 3 at 9:48
airlanggaairlangga
775
$endgroup$
This question already has an answer here:
Use any method, determine whether the series $sum_{k=1}^{infty} frac{7^{k} + k}{k! + 1}$ converges?
4 answers
$sum_{n=1}^infty frac{4^n+n}{n!}$
Using ratio test :
$lim_{n to infty} frac{a_{n+1}}{a_n}\
lim_{n to infty} frac{(4^{n+1}+n+1)n!}{(n+1)!(4^n+n)}\
lim_{n to infty} frac{(4^{n+1}+n+1)}{(n+1)(4^n+n)}$
I think it's converges, but I didn't know how to solve the limit.
This question already has an answer here:
Use any method, determine whether the series $sum_{k=1}^{infty} frac{7^{k} + k}{k! + 1}$ converges?
4 answers
calculus sequences-and-series
calculus sequences-and-series
asked Feb 3 at 9:48
airlanggaairlangga
775
asked Feb 3 at 9:48
airlanggaairlangga
775
asked Feb 3 at 9:48
airlanggaairlangga
775
asked Feb 3 at 9:48
airlanggaairlangga
775
asked Feb 3 at 9:48
airlanggaairlangga
775
775
marked as duplicate by Nosrati, Community♦ Feb 3 at 10:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Nosrati, Community♦ Feb 3 at 10:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Split ot into two series.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 3 at 9:50
add a comment |
1
$begingroup$
Split ot into two series.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 3 at 9:50
1
1
$begingroup$
Split ot into two series.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 3 at 9:50
$begingroup$
Split ot into two series.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 3 at 9:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First of all, clearly $sum frac{4^n}{n!}$ converges (to $e^4$, in fact). Now, similarly, one can use the ratio test or just notice that
$$ sum frac{n}{n!} = sum frac{1}{(n-1)!} = ... $$
answered Feb 3 at 9:51
JamesJames
2,636425
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add a comment |
$begingroup$
The function $mathrm{e}^x=sum_{n=0}frac{x^n}{n!}$ is finite for $x=4$ as well as for any other real such as $x=5$.
My approach is to show that your sum (with the $4^n+n$ in the nominator of each term) is dominated $sum_{n=0}frac{5^n}{n!}$, i.e. the number $mathrm{e}^5=148.413..$.
So what needs to be shown is just that $4^n+nle 5^n$.
Note: I start at $n=0$ here for convenience (as the exponential is defined that was), but the first term is clearly the same (namely equal to $1$) for both series and so I could also just drop it as well.
Firstly, the condition can also written as $4^nle 5^n-n$.
For $n=0$ we have that
$4^0 = 5^0 - 0$
and actually also
$4^1 = 5^1 - 1$.
Now, by induction,
$4^{(n+1)} = 4^ncdot 4le (5^n-n) cdot (5-1) = 5^{n+1} - (4,n + 5^n) le 5^{n+1}-(n+1).$
and we're done.
Note: We could even use weaker language: In principle we merely need to show that the sum is smaller than some constant, so we wouldn't have to start at $n=0$ but merely prove $4^nle 5^n-n$ for some sufficiently large $n$ onwards. That it holds for all $n$ is a simplification of the initial condition, $4^0 = 5^0 - 0$.
answered Feb 3 at 10:16
Nikolaj-KNikolaj-K
5,99223069
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First of all, clearly $sum frac{4^n}{n!}$ converges (to $e^4$, in fact). Now, similarly, one can use the ratio test or just notice that
$$ sum frac{n}{n!} = sum frac{1}{(n-1)!} = ... $$
answered Feb 3 at 9:51
JamesJames
2,636425
$endgroup$
add a comment |
$begingroup$
First of all, clearly $sum frac{4^n}{n!}$ converges (to $e^4$, in fact). Now, similarly, one can use the ratio test or just notice that
$$ sum frac{n}{n!} = sum frac{1}{(n-1)!} = ... $$
answered Feb 3 at 9:51
JamesJames
2,636425
$endgroup$
add a comment |
$begingroup$
First of all, clearly $sum frac{4^n}{n!}$ converges (to $e^4$, in fact). Now, similarly, one can use the ratio test or just notice that
$$ sum frac{n}{n!} = sum frac{1}{(n-1)!} = ... $$
answered Feb 3 at 9:51
JamesJames
2,636425
$endgroup$
First of all, clearly $sum frac{4^n}{n!}$ converges (to $e^4$, in fact). Now, similarly, one can use the ratio test or just notice that
$$ sum frac{n}{n!} = sum frac{1}{(n-1)!} = ... $$
answered Feb 3 at 9:51
JamesJames
2,636425
answered Feb 3 at 9:51
JamesJames
2,636425
answered Feb 3 at 9:51
JamesJames
2,636425
answered Feb 3 at 9:51
JamesJames
2,636425
2,636425
add a comment |
add a comment |
$begingroup$
The function $mathrm{e}^x=sum_{n=0}frac{x^n}{n!}$ is finite for $x=4$ as well as for any other real such as $x=5$.
My approach is to show that your sum (with the $4^n+n$ in the nominator of each term) is dominated $sum_{n=0}frac{5^n}{n!}$, i.e. the number $mathrm{e}^5=148.413..$.
So what needs to be shown is just that $4^n+nle 5^n$.
Note: I start at $n=0$ here for convenience (as the exponential is defined that was), but the first term is clearly the same (namely equal to $1$) for both series and so I could also just drop it as well.
Firstly, the condition can also written as $4^nle 5^n-n$.
For $n=0$ we have that
$4^0 = 5^0 - 0$
and actually also
$4^1 = 5^1 - 1$.
Now, by induction,
$4^{(n+1)} = 4^ncdot 4le (5^n-n) cdot (5-1) = 5^{n+1} - (4,n + 5^n) le 5^{n+1}-(n+1).$
and we're done.
Note: We could even use weaker language: In principle we merely need to show that the sum is smaller than some constant, so we wouldn't have to start at $n=0$ but merely prove $4^nle 5^n-n$ for some sufficiently large $n$ onwards. That it holds for all $n$ is a simplification of the initial condition, $4^0 = 5^0 - 0$.
answered Feb 3 at 10:16
Nikolaj-KNikolaj-K
5,99223069
$endgroup$
add a comment |
$begingroup$
The function $mathrm{e}^x=sum_{n=0}frac{x^n}{n!}$ is finite for $x=4$ as well as for any other real such as $x=5$.
My approach is to show that your sum (with the $4^n+n$ in the nominator of each term) is dominated $sum_{n=0}frac{5^n}{n!}$, i.e. the number $mathrm{e}^5=148.413..$.
So what needs to be shown is just that $4^n+nle 5^n$.
Note: I start at $n=0$ here for convenience (as the exponential is defined that was), but the first term is clearly the same (namely equal to $1$) for both series and so I could also just drop it as well.
Firstly, the condition can also written as $4^nle 5^n-n$.
For $n=0$ we have that
$4^0 = 5^0 - 0$
and actually also
$4^1 = 5^1 - 1$.
Now, by induction,
$4^{(n+1)} = 4^ncdot 4le (5^n-n) cdot (5-1) = 5^{n+1} - (4,n + 5^n) le 5^{n+1}-(n+1).$
and we're done.
Note: We could even use weaker language: In principle we merely need to show that the sum is smaller than some constant, so we wouldn't have to start at $n=0$ but merely prove $4^nle 5^n-n$ for some sufficiently large $n$ onwards. That it holds for all $n$ is a simplification of the initial condition, $4^0 = 5^0 - 0$.
answered Feb 3 at 10:16
Nikolaj-KNikolaj-K
5,99223069
$endgroup$
add a comment |
$begingroup$
The function $mathrm{e}^x=sum_{n=0}frac{x^n}{n!}$ is finite for $x=4$ as well as for any other real such as $x=5$.
My approach is to show that your sum (with the $4^n+n$ in the nominator of each term) is dominated $sum_{n=0}frac{5^n}{n!}$, i.e. the number $mathrm{e}^5=148.413..$.
So what needs to be shown is just that $4^n+nle 5^n$.
Note: I start at $n=0$ here for convenience (as the exponential is defined that was), but the first term is clearly the same (namely equal to $1$) for both series and so I could also just drop it as well.
Firstly, the condition can also written as $4^nle 5^n-n$.
For $n=0$ we have that
$4^0 = 5^0 - 0$
and actually also
$4^1 = 5^1 - 1$.
Now, by induction,
$4^{(n+1)} = 4^ncdot 4le (5^n-n) cdot (5-1) = 5^{n+1} - (4,n + 5^n) le 5^{n+1}-(n+1).$
and we're done.
Note: We could even use weaker language: In principle we merely need to show that the sum is smaller than some constant, so we wouldn't have to start at $n=0$ but merely prove $4^nle 5^n-n$ for some sufficiently large $n$ onwards. That it holds for all $n$ is a simplification of the initial condition, $4^0 = 5^0 - 0$.
answered Feb 3 at 10:16
Nikolaj-KNikolaj-K
5,99223069
$endgroup$
The function $mathrm{e}^x=sum_{n=0}frac{x^n}{n!}$ is finite for $x=4$ as well as for any other real such as $x=5$.
My approach is to show that your sum (with the $4^n+n$ in the nominator of each term) is dominated $sum_{n=0}frac{5^n}{n!}$, i.e. the number $mathrm{e}^5=148.413..$.
So what needs to be shown is just that $4^n+nle 5^n$.
Note: I start at $n=0$ here for convenience (as the exponential is defined that was), but the first term is clearly the same (namely equal to $1$) for both series and so I could also just drop it as well.
Firstly, the condition can also written as $4^nle 5^n-n$.
For $n=0$ we have that
$4^0 = 5^0 - 0$
and actually also
$4^1 = 5^1 - 1$.
Now, by induction,
$4^{(n+1)} = 4^ncdot 4le (5^n-n) cdot (5-1) = 5^{n+1} - (4,n + 5^n) le 5^{n+1}-(n+1).$
and we're done.
Note: We could even use weaker language: In principle we merely need to show that the sum is smaller than some constant, so we wouldn't have to start at $n=0$ but merely prove $4^nle 5^n-n$ for some sufficiently large $n$ onwards. That it holds for all $n$ is a simplification of the initial condition, $4^0 = 5^0 - 0$.
answered Feb 3 at 10:16
Nikolaj-KNikolaj-K
5,99223069
edited Feb 3 at 10:33
answered Feb 3 at 10:16
Nikolaj-KNikolaj-K
5,99223069
answered Feb 3 at 10:16
Nikolaj-KNikolaj-K
5,99223069
answered Feb 3 at 10:16
Nikolaj-KNikolaj-K
5,99223069
5,99223069
add a comment |
add a comment |
1
$begingroup$
Split ot into two series.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 3 at 9:50