Let $S_n = \sum_{i = 1}^n \frac{i^k}{n^{k+1}} $ . For what values of $k$ the series $S_n$ is convergent and what is the value of convergence ? I'm really unable to understand $S_n$ because I haven't seen any series which is similar to that , so applying tests isn't possible for me . Also I've searched over the internet but didn't find any useful result .
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@gimusi When $-1 \lt k \le 0$ , $S_n$ converges or diverges ? – S.H.W Mar 20 '18 at 14:38
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my hint was given for the case $k \in \mathbb{Z^+}$ since I thought for this limitation, you can refer to the other answer for the general case (as for example by the integral test). – user Mar 20 '18 at 15:24
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Possibly the OP has a different problem in mind than what was treated in the proposed duplicate. Note the wording "For what values of $k$ the series $S_n$ is convergent?" Since $S_n$ is itself a finite series, it trivially converges. Perhaps the issue is whether the sequence ${S_n}$ converges, as $n\to \infty$, the topic addressed by the proposed duplicate (at least for some exponents $m$). The Comment left by the OP above asks about exponents in the range $-1\lt k \le 0$, where $k$ in this Question plays the role of $m$ in the proposed duplicate. Voting to close as unclear. – hardmath Sep 21 '18 at 18:10
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Hint
- If $k>0$, $$S_n=\frac{1}{n}\sum_{i=1}^n\left(\frac{i}{n}\right)^k\underset{n\to \infty }{\longrightarrow} \int_0^1 x^kdx.$$
I let you see the case where $k\leq 0$.
If $k=0$ it's obvious (since $(S_n)$ is constant).
If $k<0$ you can easily show that $(S_n)$ diverge.
Surb
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So by using improper integrals $S_n$ converges when $k \gt -1$ and convergence value is $\frac{1}{k+1}$ ? – S.H.W Mar 18 '18 at 23:37
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@S.H.W : I'm sure you can answer this question look that $S_n$ is something like $n^{k-1}\sum_{i=1}^n \frac{1}{i^k}$. – Surb Mar 19 '18 at 10:41
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@Surb My teacher told me that it converges when $-1 \lt k \le 0 $ . Is it wrong ? – S.H.W Mar 20 '18 at 05:29
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Since $x^k$ monotonically increases, we have
$$\frac{n^{k+1}-1}{k+1}=\int_1^{n} x^k\,dx\le \sum_{k=1}^n i^k\le \int_1^{n+1}x^k\,dx=\frac{(n+1)^{k+1}}{k+1}\tag 1$$
whence dividing by $n^{k+1}$, letting $n\to\infty$, and applying the squeeze theorem to $(1)$, we find that
$$\lim_{n\to \infty}\sum_{i=1}^n\frac{i^k}{n^{k+1}}=\frac1{k+1}$$
Mark Viola
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user
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@S.H.W It is a general result which can be proved by induction or by Faulhaber's formula (I've just added the link). – user Mar 18 '18 at 23:26
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If you enjoy generalized harmonic numbers, you can find the limit and approximate the partial sums since $$S_n = \sum_{i = 1}^n \frac{i^k}{n^{k+1}}=\frac{1}{n^{k+1}}\sum_{i = 1}^n i^k=\frac{H_n^{(-k)}}{n^{k+1}}$$ Now, using asymptotics $$H_n^{(-k)}=n^k \left(\frac{n}{k+1}+\frac{1}{2}+\frac{k}{12 n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-k)$$ and get $$S_n=\left(\frac{1}{k+1}+\frac{1}{2 n}+\frac{k}{12 n^2}+O\left(\frac{1}{n^4}\right)\right)+\frac {\zeta (-k)}{n^{k+1}}$$ For illustration purposes, using $k=\pi$, here are a few values $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 1 & 1.0106954353 & 1.0000000000 \\ 2 & 0.5573245570 & 0.5566573661 \\ 3 & 0.4372871463 & 0.4371554062 \\ 4 & 0.3828393613 & 0.3827976837 \\ 5 & 0.3519344645 & 0.3519173945 \\ 6 & 0.3320630018 & 0.3320547700 \\ 7 & 0.3182267765 & 0.3182223333 \\ 8 & 0.3080449761 & 0.3080423716 \\ 9 & 0.3002414849 & 0.3002398589 \\ 10 & 0.2940715381 & 0.2940704713 \\ 11 & 0.2890715458 & 0.2890708172 \\ 12 & 0.2849379775 & 0.2849374630 \\ 13 & 0.2814638355 & 0.2814634620 \\ 14 & 0.2785031372 & 0.2785028595 \\ 15 & 0.2759499934 & 0.2759497827 \\ 16 & 0.2737257376 & 0.2737255748 \\ 17 & 0.2717707116 & 0.2717705839 \\ 18 & 0.2700388547 & 0.2700387531 \\ 19 & 0.2684940402 & 0.2684939583 \\ 20 & 0.2671075359 & 0.2671074692 \end{array} \right)$$
Claude Leibovici
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