Mixing
How to evaluate: $limlimits_{ntoinfty} frac{1^{p-1}+2^{p-1}+…+n^{p-1}}{n^p}$
$begingroup$
How to evaluate: $$lim_{ntoinfty} dfrac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}$$
when
$i)$ $pinmathbb R,pneq0$
$ii)space p=0$
So for $i)$ I tried using Stolz–Cesàro theorem and Binomial theorem and If I didn't mess up I got $1$. But I'm unsure about it, but for $ii)$ and I don't have a clue where to begin with.
calculus limits analysis limits-without-lhopital
edited Feb 3 at 11:14
Martin Sleziak
45k10123277
asked Jan 25 at 10:30
iggykimiiggykimi
31910
$endgroup$
add a comment |
$begingroup$
How to evaluate: $$lim_{ntoinfty} dfrac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}$$
when
$i)$ $pinmathbb R,pneq0$
$ii)space p=0$
So for $i)$ I tried using Stolz–Cesàro theorem and Binomial theorem and If I didn't mess up I got $1$. But I'm unsure about it, but for $ii)$ and I don't have a clue where to begin with.
calculus limits analysis limits-without-lhopital
edited Feb 3 at 11:14
Martin Sleziak
45k10123277
asked Jan 25 at 10:30
iggykimiiggykimi
31910
$endgroup$
$begingroup$
I would use series / integral comparison to bound below and above $1^{p-1}+2^{p-1}+…+n^{p-1}$.
$endgroup$
– mathcounterexamples.net
Jan 25 at 10:33
$begingroup$
Use math.stackexchange.com/questions/469885/…
$endgroup$
– lab bhattacharjee
Jan 25 at 10:34
$begingroup$
$$lim_{ntoinfty}frac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}=lim_{ntoinfty}sum_{j=1}^nleft(frac jnright)^{p-1}frac1n=int_0^1x^{p-1}dx=begin{cases}frac1p,&pne0\infty,&p=0end{cases}$$
$endgroup$
– Shubham Johri
Jan 25 at 10:37
1
$begingroup$
For $p=0$, the limit is the sum of the Harmonic series ${frac1n}$. Proofs of its divergence are quite popular; see here
$endgroup$
– Shubham Johri
Jan 25 at 10:40
$begingroup$
For $p>0$ you can have a look at Evaluate $limlimits_{ntoinfty}frac{sum_{k=1}^n k^m}{n^{m+1}}$ (and the questions linked there) or What is the result of $ lim_{n to infty} frac{ sum^n_{i=1} i^k}{n^{k+1}}, k in mathbb{R} $ and why? (and the questions linked there).
$endgroup$
– Martin Sleziak
Feb 3 at 11:21
add a comment |
$begingroup$
How to evaluate: $$lim_{ntoinfty} dfrac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}$$
when
$i)$ $pinmathbb R,pneq0$
$ii)space p=0$
So for $i)$ I tried using Stolz–Cesàro theorem and Binomial theorem and If I didn't mess up I got $1$. But I'm unsure about it, but for $ii)$ and I don't have a clue where to begin with.
calculus limits analysis limits-without-lhopital
edited Feb 3 at 11:14
Martin Sleziak
45k10123277
asked Jan 25 at 10:30
iggykimiiggykimi
31910
$endgroup$
How to evaluate: $$lim_{ntoinfty} dfrac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}$$
when
$i)$ $pinmathbb R,pneq0$
$ii)space p=0$
So for $i)$ I tried using Stolz–Cesàro theorem and Binomial theorem and If I didn't mess up I got $1$. But I'm unsure about it, but for $ii)$ and I don't have a clue where to begin with.
calculus limits analysis limits-without-lhopital
calculus limits analysis limits-without-lhopital
edited Feb 3 at 11:14
Martin Sleziak
45k10123277
asked Jan 25 at 10:30
iggykimiiggykimi
31910
edited Feb 3 at 11:14
Martin Sleziak
45k10123277
asked Jan 25 at 10:30
iggykimiiggykimi
31910
edited Feb 3 at 11:14
Martin Sleziak
45k10123277
edited Feb 3 at 11:14
Martin Sleziak
45k10123277
edited Feb 3 at 11:14
Martin Sleziak
45k10123277
45k10123277
asked Jan 25 at 10:30
iggykimiiggykimi
31910
asked Jan 25 at 10:30
iggykimiiggykimi
31910
asked Jan 25 at 10:30
iggykimiiggykimi
31910
31910
$begingroup$
I would use series / integral comparison to bound below and above $1^{p-1}+2^{p-1}+…+n^{p-1}$.
$endgroup$
– mathcounterexamples.net
Jan 25 at 10:33
$begingroup$
Use math.stackexchange.com/questions/469885/…
$endgroup$
– lab bhattacharjee
Jan 25 at 10:34
$begingroup$
$$lim_{ntoinfty}frac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}=lim_{ntoinfty}sum_{j=1}^nleft(frac jnright)^{p-1}frac1n=int_0^1x^{p-1}dx=begin{cases}frac1p,&pne0\infty,&p=0end{cases}$$
$endgroup$
– Shubham Johri
Jan 25 at 10:37
1
$begingroup$
For $p=0$, the limit is the sum of the Harmonic series ${frac1n}$. Proofs of its divergence are quite popular; see here
$endgroup$
– Shubham Johri
Jan 25 at 10:40
$begingroup$
For $p>0$ you can have a look at Evaluate $limlimits_{ntoinfty}frac{sum_{k=1}^n k^m}{n^{m+1}}$ (and the questions linked there) or What is the result of $ lim_{n to infty} frac{ sum^n_{i=1} i^k}{n^{k+1}}, k in mathbb{R} $ and why? (and the questions linked there).
$endgroup$
– Martin Sleziak
Feb 3 at 11:21
add a comment |
$begingroup$
I would use series / integral comparison to bound below and above $1^{p-1}+2^{p-1}+…+n^{p-1}$.
$endgroup$
– mathcounterexamples.net
Jan 25 at 10:33
$begingroup$
Use math.stackexchange.com/questions/469885/…
$endgroup$
– lab bhattacharjee
Jan 25 at 10:34
$begingroup$
$$lim_{ntoinfty}frac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}=lim_{ntoinfty}sum_{j=1}^nleft(frac jnright)^{p-1}frac1n=int_0^1x^{p-1}dx=begin{cases}frac1p,&pne0\infty,&p=0end{cases}$$
$endgroup$
– Shubham Johri
Jan 25 at 10:37
1
$begingroup$
For $p=0$, the limit is the sum of the Harmonic series ${frac1n}$. Proofs of its divergence are quite popular; see here
$endgroup$
– Shubham Johri
Jan 25 at 10:40
$begingroup$
For $p>0$ you can have a look at Evaluate $limlimits_{ntoinfty}frac{sum_{k=1}^n k^m}{n^{m+1}}$ (and the questions linked there) or What is the result of $ lim_{n to infty} frac{ sum^n_{i=1} i^k}{n^{k+1}}, k in mathbb{R} $ and why? (and the questions linked there).
$endgroup$
– Martin Sleziak
Feb 3 at 11:21
$begingroup$
I would use series / integral comparison to bound below and above $1^{p-1}+2^{p-1}+…+n^{p-1}$.
$endgroup$
– mathcounterexamples.net
Jan 25 at 10:33
$begingroup$
I would use series / integral comparison to bound below and above $1^{p-1}+2^{p-1}+…+n^{p-1}$.
$endgroup$
– mathcounterexamples.net
Jan 25 at 10:33
$begingroup$
Use math.stackexchange.com/questions/469885/…
$endgroup$
– lab bhattacharjee
Jan 25 at 10:34
$begingroup$
Use math.stackexchange.com/questions/469885/…
$endgroup$
– lab bhattacharjee
Jan 25 at 10:34
$begingroup$
$$lim_{ntoinfty}frac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}=lim_{ntoinfty}sum_{j=1}^nleft(frac jnright)^{p-1}frac1n=int_0^1x^{p-1}dx=begin{cases}frac1p,&pne0\infty,&p=0end{cases}$$
$endgroup$
– Shubham Johri
Jan 25 at 10:37
$begingroup$
$$lim_{ntoinfty}frac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}=lim_{ntoinfty}sum_{j=1}^nleft(frac jnright)^{p-1}frac1n=int_0^1x^{p-1}dx=begin{cases}frac1p,&pne0\infty,&p=0end{cases}$$
$endgroup$
– Shubham Johri
Jan 25 at 10:37
1
1
$begingroup$
For $p=0$, the limit is the sum of the Harmonic series ${frac1n}$. Proofs of its divergence are quite popular; see here
$endgroup$
– Shubham Johri
Jan 25 at 10:40
$begingroup$
For $p=0$, the limit is the sum of the Harmonic series ${frac1n}$. Proofs of its divergence are quite popular; see here
$endgroup$
– Shubham Johri
Jan 25 at 10:40
$begingroup$
For $p>0$ you can have a look at Evaluate $limlimits_{ntoinfty}frac{sum_{k=1}^n k^m}{n^{m+1}}$ (and the questions linked there) or What is the result of $ lim_{n to infty} frac{ sum^n_{i=1} i^k}{n^{k+1}}, k in mathbb{R} $ and why? (and the questions linked there).
$endgroup$
– Martin Sleziak
Feb 3 at 11:21
$begingroup$
For $p>0$ you can have a look at Evaluate $limlimits_{ntoinfty}frac{sum_{k=1}^n k^m}{n^{m+1}}$ (and the questions linked there) or What is the result of $ lim_{n to infty} frac{ sum^n_{i=1} i^k}{n^{k+1}}, k in mathbb{R} $ and why? (and the questions linked there).
$endgroup$
– Martin Sleziak
Feb 3 at 11:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We can apply the Stolz-Cesàro theorem for positive real values $p, pne 1$
and consider other values of $p$ separately.
We consider for $pinmathbb{R}$:
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}tag{1}
end{align*}
Case $p>1, 0<p<1$:
If $p>1$ resp. $0<p<1$ the sequence $(n^p)_{ngeq 1}$ is strictly monotone increasing and unbounded. We can apply the Stolz-Cesàro theorem by letting
begin{align*}
a_n&=1^{p-1}+2^{p-1}+cdots+n^{p-1}\
b_n&=n^p=underbrace{n^{p-1}+n^{p-1}+cdots+n^{p-1}}_{n mathrm{ times}}
end{align*}
Since
begin{align*}
lim_{nto infty}frac{a_n-a_{n-1}}{b_n-b_{n-1}}&=lim_{ntoinfty}frac{n^{p-1}}{n^p-(n-1)^{p}}\
&=lim_{ntoinfty}frac{n^{p-1}}{n^p-(n^p-pn^{p-1}+binom{p}{2}n^{p-2}-cdots)}\
&=lim_{ntoinfty}frac{n^{p-1}}{pn^{p-1}-binom{p}{2}n^{p-2}+cdots}\
&=frac{1}{p}
end{align*}
we have according to the theorem
begin{align*}
lim_{nto infty}frac{a_n}{b_n}&=lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}color{blue}{=frac{1}{p}}
end{align*}
Case $p=1$:
We obtain
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}
=lim_{ntoinfty}frac{overbrace{1+1+cdots+1}^{nmathrm{ times}}}{n}=lim_{ntoinfty}frac{n}{n}
color{blue}{=1}
end{align*}
Case $p=0$:
We obtain
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}
=lim_{ntoinfty}left(1+frac{1}{2}+cdots+frac{1}{n}right)=sum_{n=1}^infty frac{1}{n}color{blue}{=infty}
end{align*}
since the harmonic series is divergent.
Case $p<0$:
We set $q:=-p$ and obtain with $q>0$:
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}
=lim_{nto infty}n^qleft(1+frac{1}{2^{q+1}}+cdots+frac{1}{n^{q+1}}right)geq lim_{ntoinfty} n^q
color{blue}{=infty}
end{align*}
We summarize:
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}=
begin{cases}
frac{1}{p}&p>0\
infty&pleq 0
end{cases}
end{align*}
answered Jan 26 at 12:02
Markus ScheuerMarkus Scheuer
64.5k460152
$endgroup$
add a comment |
$begingroup$
Hint:
Just write
$$dfrac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p} = frac{1}{n}sum_{k=1}^nleft(frac{k}{n}right)^{p-1}$$
and handle it as the limit of a Riemann sum.
answered Jan 25 at 10:39
trancelocationtrancelocation
14.2k1829
$endgroup$
$begingroup$
Sorry but I haven't learnt about integration yet and I don't know how to do that :(
$endgroup$
– iggykimi
Jan 25 at 18:10
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We can apply the Stolz-Cesàro theorem for positive real values $p, pne 1$
and consider other values of $p$ separately.
We consider for $pinmathbb{R}$:
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}tag{1}
end{align*}
Case $p>1, 0<p<1$:
If $p>1$ resp. $0<p<1$ the sequence $(n^p)_{ngeq 1}$ is strictly monotone increasing and unbounded. We can apply the Stolz-Cesàro theorem by letting
begin{align*}
a_n&=1^{p-1}+2^{p-1}+cdots+n^{p-1}\
b_n&=n^p=underbrace{n^{p-1}+n^{p-1}+cdots+n^{p-1}}_{n mathrm{ times}}
end{align*}
Since
begin{align*}
lim_{nto infty}frac{a_n-a_{n-1}}{b_n-b_{n-1}}&=lim_{ntoinfty}frac{n^{p-1}}{n^p-(n-1)^{p}}\
&=lim_{ntoinfty}frac{n^{p-1}}{n^p-(n^p-pn^{p-1}+binom{p}{2}n^{p-2}-cdots)}\
&=lim_{ntoinfty}frac{n^{p-1}}{pn^{p-1}-binom{p}{2}n^{p-2}+cdots}\
&=frac{1}{p}
end{align*}
we have according to the theorem
begin{align*}
lim_{nto infty}frac{a_n}{b_n}&=lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}color{blue}{=frac{1}{p}}
end{align*}
Case $p=1$:
We obtain
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}
=lim_{ntoinfty}frac{overbrace{1+1+cdots+1}^{nmathrm{ times}}}{n}=lim_{ntoinfty}frac{n}{n}
color{blue}{=1}
end{align*}
Case $p=0$:
We obtain
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}
=lim_{ntoinfty}left(1+frac{1}{2}+cdots+frac{1}{n}right)=sum_{n=1}^infty frac{1}{n}color{blue}{=infty}
end{align*}
since the harmonic series is divergent.
Case $p<0$:
We set $q:=-p$ and obtain with $q>0$:
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}
=lim_{nto infty}n^qleft(1+frac{1}{2^{q+1}}+cdots+frac{1}{n^{q+1}}right)geq lim_{ntoinfty} n^q
color{blue}{=infty}
end{align*}
We summarize:
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}=
begin{cases}
frac{1}{p}&p>0\
infty&pleq 0
end{cases}
end{align*}
answered Jan 26 at 12:02
Markus ScheuerMarkus Scheuer
64.5k460152
$endgroup$
add a comment |
$begingroup$
We can apply the Stolz-Cesàro theorem for positive real values $p, pne 1$
and consider other values of $p$ separately.
We consider for $pinmathbb{R}$:
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}tag{1}
end{align*}
Case $p>1, 0<p<1$:
If $p>1$ resp. $0<p<1$ the sequence $(n^p)_{ngeq 1}$ is strictly monotone increasing and unbounded. We can apply the Stolz-Cesàro theorem by letting
begin{align*}
a_n&=1^{p-1}+2^{p-1}+cdots+n^{p-1}\
b_n&=n^p=underbrace{n^{p-1}+n^{p-1}+cdots+n^{p-1}}_{n mathrm{ times}}
end{align*}
Since
begin{align*}
lim_{nto infty}frac{a_n-a_{n-1}}{b_n-b_{n-1}}&=lim_{ntoinfty}frac{n^{p-1}}{n^p-(n-1)^{p}}\
&=lim_{ntoinfty}frac{n^{p-1}}{n^p-(n^p-pn^{p-1}+binom{p}{2}n^{p-2}-cdots)}\
&=lim_{ntoinfty}frac{n^{p-1}}{pn^{p-1}-binom{p}{2}n^{p-2}+cdots}\
&=frac{1}{p}
end{align*}
we have according to the theorem
begin{align*}
lim_{nto infty}frac{a_n}{b_n}&=lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}color{blue}{=frac{1}{p}}
end{align*}
Case $p=1$:
We obtain
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}
=lim_{ntoinfty}frac{overbrace{1+1+cdots+1}^{nmathrm{ times}}}{n}=lim_{ntoinfty}frac{n}{n}
color{blue}{=1}
end{align*}
Case $p=0$:
We obtain
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}
=lim_{ntoinfty}left(1+frac{1}{2}+cdots+frac{1}{n}right)=sum_{n=1}^infty frac{1}{n}color{blue}{=infty}
end{align*}
since the harmonic series is divergent.
Case $p<0$:
We set $q:=-p$ and obtain with $q>0$:
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}
=lim_{nto infty}n^qleft(1+frac{1}{2^{q+1}}+cdots+frac{1}{n^{q+1}}right)geq lim_{ntoinfty} n^q
color{blue}{=infty}
end{align*}
We summarize:
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}=
begin{cases}
frac{1}{p}&p>0\
infty&pleq 0
end{cases}
end{align*}
answered Jan 26 at 12:02
Markus ScheuerMarkus Scheuer
64.5k460152
$endgroup$
add a comment |
$begingroup$
We can apply the Stolz-Cesàro theorem for positive real values $p, pne 1$
and consider other values of $p$ separately.
We consider for $pinmathbb{R}$:
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}tag{1}
end{align*}
Case $p>1, 0<p<1$:
If $p>1$ resp. $0<p<1$ the sequence $(n^p)_{ngeq 1}$ is strictly monotone increasing and unbounded. We can apply the Stolz-Cesàro theorem by letting
begin{align*}
a_n&=1^{p-1}+2^{p-1}+cdots+n^{p-1}\
b_n&=n^p=underbrace{n^{p-1}+n^{p-1}+cdots+n^{p-1}}_{n mathrm{ times}}
end{align*}
Since
begin{align*}
lim_{nto infty}frac{a_n-a_{n-1}}{b_n-b_{n-1}}&=lim_{ntoinfty}frac{n^{p-1}}{n^p-(n-1)^{p}}\
&=lim_{ntoinfty}frac{n^{p-1}}{n^p-(n^p-pn^{p-1}+binom{p}{2}n^{p-2}-cdots)}\
&=lim_{ntoinfty}frac{n^{p-1}}{pn^{p-1}-binom{p}{2}n^{p-2}+cdots}\
&=frac{1}{p}
end{align*}
we have according to the theorem
begin{align*}
lim_{nto infty}frac{a_n}{b_n}&=lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}color{blue}{=frac{1}{p}}
end{align*}
Case $p=1$:
We obtain
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}
=lim_{ntoinfty}frac{overbrace{1+1+cdots+1}^{nmathrm{ times}}}{n}=lim_{ntoinfty}frac{n}{n}
color{blue}{=1}
end{align*}
Case $p=0$:
We obtain
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}
=lim_{ntoinfty}left(1+frac{1}{2}+cdots+frac{1}{n}right)=sum_{n=1}^infty frac{1}{n}color{blue}{=infty}
end{align*}
since the harmonic series is divergent.
Case $p<0$:
We set $q:=-p$ and obtain with $q>0$:
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}
=lim_{nto infty}n^qleft(1+frac{1}{2^{q+1}}+cdots+frac{1}{n^{q+1}}right)geq lim_{ntoinfty} n^q
color{blue}{=infty}
end{align*}
We summarize:
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}=
begin{cases}
frac{1}{p}&p>0\
infty&pleq 0
end{cases}
end{align*}
answered Jan 26 at 12:02
Markus ScheuerMarkus Scheuer
64.5k460152
$endgroup$
We can apply the Stolz-Cesàro theorem for positive real values $p, pne 1$
and consider other values of $p$ separately.
We consider for $pinmathbb{R}$:
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}tag{1}
end{align*}
Case $p>1, 0<p<1$:
If $p>1$ resp. $0<p<1$ the sequence $(n^p)_{ngeq 1}$ is strictly monotone increasing and unbounded. We can apply the Stolz-Cesàro theorem by letting
begin{align*}
a_n&=1^{p-1}+2^{p-1}+cdots+n^{p-1}\
b_n&=n^p=underbrace{n^{p-1}+n^{p-1}+cdots+n^{p-1}}_{n mathrm{ times}}
end{align*}
Since
begin{align*}
lim_{nto infty}frac{a_n-a_{n-1}}{b_n-b_{n-1}}&=lim_{ntoinfty}frac{n^{p-1}}{n^p-(n-1)^{p}}\
&=lim_{ntoinfty}frac{n^{p-1}}{n^p-(n^p-pn^{p-1}+binom{p}{2}n^{p-2}-cdots)}\
&=lim_{ntoinfty}frac{n^{p-1}}{pn^{p-1}-binom{p}{2}n^{p-2}+cdots}\
&=frac{1}{p}
end{align*}
we have according to the theorem
begin{align*}
lim_{nto infty}frac{a_n}{b_n}&=lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}color{blue}{=frac{1}{p}}
end{align*}
Case $p=1$:
We obtain
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}
=lim_{ntoinfty}frac{overbrace{1+1+cdots+1}^{nmathrm{ times}}}{n}=lim_{ntoinfty}frac{n}{n}
color{blue}{=1}
end{align*}
Case $p=0$:
We obtain
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}
=lim_{ntoinfty}left(1+frac{1}{2}+cdots+frac{1}{n}right)=sum_{n=1}^infty frac{1}{n}color{blue}{=infty}
end{align*}
since the harmonic series is divergent.
Case $p<0$:
We set $q:=-p$ and obtain with $q>0$:
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}
=lim_{nto infty}n^qleft(1+frac{1}{2^{q+1}}+cdots+frac{1}{n^{q+1}}right)geq lim_{ntoinfty} n^q
color{blue}{=infty}
end{align*}
We summarize:
begin{align*}
lim_{nto infty}frac{1^{p-1}+2^{p-1}+cdots+n^{p-1}}{n^p}=
begin{cases}
frac{1}{p}&p>0\
infty&pleq 0
end{cases}
end{align*}
answered Jan 26 at 12:02
Markus ScheuerMarkus Scheuer
64.5k460152
answered Jan 26 at 12:02
Markus ScheuerMarkus Scheuer
64.5k460152
answered Jan 26 at 12:02
Markus ScheuerMarkus Scheuer
64.5k460152
answered Jan 26 at 12:02
Markus ScheuerMarkus Scheuer
64.5k460152
64.5k460152
add a comment |
add a comment |
$begingroup$
Hint:
Just write
$$dfrac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p} = frac{1}{n}sum_{k=1}^nleft(frac{k}{n}right)^{p-1}$$
and handle it as the limit of a Riemann sum.
answered Jan 25 at 10:39
trancelocationtrancelocation
14.2k1829
$endgroup$
$begingroup$
Sorry but I haven't learnt about integration yet and I don't know how to do that :(
$endgroup$
– iggykimi
Jan 25 at 18:10
add a comment |
$begingroup$
Hint:
Just write
$$dfrac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p} = frac{1}{n}sum_{k=1}^nleft(frac{k}{n}right)^{p-1}$$
and handle it as the limit of a Riemann sum.
answered Jan 25 at 10:39
trancelocationtrancelocation
14.2k1829
$endgroup$
$begingroup$
Sorry but I haven't learnt about integration yet and I don't know how to do that :(
$endgroup$
– iggykimi
Jan 25 at 18:10
add a comment |
$begingroup$
Hint:
Just write
$$dfrac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p} = frac{1}{n}sum_{k=1}^nleft(frac{k}{n}right)^{p-1}$$
and handle it as the limit of a Riemann sum.
answered Jan 25 at 10:39
trancelocationtrancelocation
14.2k1829
$endgroup$
Hint:
Just write
$$dfrac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p} = frac{1}{n}sum_{k=1}^nleft(frac{k}{n}right)^{p-1}$$
and handle it as the limit of a Riemann sum.
answered Jan 25 at 10:39
trancelocationtrancelocation
14.2k1829
answered Jan 25 at 10:39
trancelocationtrancelocation
14.2k1829
answered Jan 25 at 10:39
trancelocationtrancelocation
14.2k1829
answered Jan 25 at 10:39
trancelocationtrancelocation
14.2k1829
14.2k1829
$begingroup$
Sorry but I haven't learnt about integration yet and I don't know how to do that :(
$endgroup$
– iggykimi
Jan 25 at 18:10
add a comment |
$begingroup$
Sorry but I haven't learnt about integration yet and I don't know how to do that :(
$endgroup$
– iggykimi
Jan 25 at 18:10
$begingroup$
Sorry but I haven't learnt about integration yet and I don't know how to do that :(
$endgroup$
– iggykimi
Jan 25 at 18:10
$begingroup$
Sorry but I haven't learnt about integration yet and I don't know how to do that :(
$endgroup$
– iggykimi
Jan 25 at 18:10
add a comment |
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$begingroup$
I would use series / integral comparison to bound below and above $1^{p-1}+2^{p-1}+…+n^{p-1}$.
$endgroup$
– mathcounterexamples.net
Jan 25 at 10:33
$begingroup$
Use math.stackexchange.com/questions/469885/…
$endgroup$
– lab bhattacharjee
Jan 25 at 10:34
$begingroup$
$$lim_{ntoinfty}frac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}=lim_{ntoinfty}sum_{j=1}^nleft(frac jnright)^{p-1}frac1n=int_0^1x^{p-1}dx=begin{cases}frac1p,&pne0\infty,&p=0end{cases}$$
$endgroup$
– Shubham Johri
Jan 25 at 10:37
1
$begingroup$
For $p=0$, the limit is the sum of the Harmonic series ${frac1n}$. Proofs of its divergence are quite popular; see here
$endgroup$
– Shubham Johri
Jan 25 at 10:40
$begingroup$
For $p>0$ you can have a look at Evaluate $limlimits_{ntoinfty}frac{sum_{k=1}^n k^m}{n^{m+1}}$ (and the questions linked there) or What is the result of $ lim_{n to infty} frac{ sum^n_{i=1} i^k}{n^{k+1}}, k in mathbb{R} $ and why? (and the questions linked there).
$endgroup$
– Martin Sleziak
Feb 3 at 11:21