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Let $k \in \mathbb N \setminus \{ 0 \}$. Find the limit $$\lim \bigg ( \frac 1 {n^k} \sum_{j = 1}^n j^k - \frac n {k+1} \bigg )$$

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    $\begingroup$ If it helps: this looks like integration numerics to me: if $f(x)=x^k$, your term is equal to $$n\left[\frac{1}{n}\sum_{j=1}^n f(\frac{j}{n})-\int_0^1f(t)dt\right].$$ In other words, you are looking for an equivalent of the error in the method of rectangles. $\endgroup$ – John Steinbeck Jan 23 '16 at 18:47
  • $\begingroup$ What is $\mathbb{N}^*$? $\endgroup$ – Eric Auld Jan 23 '16 at 18:57
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Ok, using my comment:

$$n\left[\frac{1}{n}\sum_{j=1}^nf(\frac{j}{n})-\int_0^1f(t)dt\right]=-n\sum_j\int_{\frac{j-1}{n}}^{\frac{j}{n}}(f(t)-f(\frac{j}{n}))dt$$ $$=-n\sum_j\int_{\frac{j-1}{n}}^{\frac{j}{n}}\left(f'\left(\frac{j}{n}\right)\left(t-\frac{j}{n}\right)+f''(c_j)\left(t-\frac{j}{n}\right)^2\right)dt$$ The term $$-n\sum_j\int_{\frac{j-1}{n}}^{\frac{j}{n}}f''(c_j)\left(t-\frac{j}{n}\right)^2dt$$ clearly goes to 0 as $n$ goes to $\infty$ (using for example that $f''$ is bounded). Therefore you only focus on $$-n\sum_j\int_{\frac{j-1}{n}}^{\frac{j}{n}}f'\left(\frac{j}{n}\right)\left(t-\frac{j}{n}\right)dt=n\sum_jf'\left(\frac{j}{n}\right)\frac{1}{2n^2}$$ whose limit is $$\frac{1}{2}\int_0^1f'(t)dt$$ which is, in your case, $$\frac{1}{2}\int_0^1kt^{k-1}dt=\frac{1}{2}$$

Hoping there is no mistake, though a good point is that for $k=1$, you easily check that your sequence is constant equal to $1/2$.

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Using Bernoulli's formula

$$\sum \limits _{j=1} ^n j^k = \frac 1 {k+1}\sum \limits _ {j=0} ^k {k+1\choose j} B_j (n+1)^{k+1-j}$$

and writing $\sum \limits _{j=0} ^k$ as the sum of the term with $j=0$, the term with $j=1$ and the rest of them, it is easy to note that your limit is

$$\lim \left( \frac 1 {k+1} B_0 \frac {(n+1)^{k+1}} {n^k} - \frac n {k+1} \right) + \lim B_1 \frac {(n+1)^k} {n^k} + \lim \frac 1 {n^k} \left( \frac 1 {k+1} \sum \limits _{j=2} ^k \binom {k+1} j B_j (n+1)^{k+1-j} \right) .$$

Since $B_0 = 1$, the first term is

$$\frac 1 {k+1} \lim \frac {(n+1)^{k+1} - n^{k+1}} {n^k} = \\ \frac 1 {k+1} \lim \frac {n ^{k+1} +(k+1) n^k + (\text{a finite sum of terms of smaller degree in } n) - n^{k+1}} {n^k} = 1 .$$

The second term is $B_1 = - \frac 1 2$ (look into the link given at the beginning of my answer to see the recurrence formula for Bernouli's numbers).

The third term is $0$, because the parantheses contain a finite sum of powers of $n$ smaller than the $k$-th power, so when you divide this by $n^k$ and take the limit the result is $0$.

Therefore, the result is $1 - \frac 1 2 + 0 = \color{blue} {\frac 1 2}$.

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