# Find the limit involving a Riemann sum.

Evaluate the two limits.

$$\lim_{n\to \infty} \left(\frac1n\right)^{15}\sum_{k=1}^nk^{15} \tag{1}$$ $$\lim_{n\to \infty} \left(\frac1n\right)^{17}\sum_{k=1}^nk^{15} \tag{2}$$

can anyone please help me with them? I know that I should use Riemann sum, but I'm not sure how or what function's Riemann integral looks this way.

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Remember that the general term in the Riemann summation is $\ f(a \ + \ i \cdot \frac{b-a}{n} ) \ \cdot \ \frac{b-a}{n} \$ (since we use "right-endpoint rule" in constructing these limits). We see that $\ b - a = 1 \ ,$ but we don't see any "a" term in the "function factor". So it must be that $\ a = 0 \$ and the rest of the definite integral expression follows from there. (This is a very common textbook and exam problem, so expect to see it again in your future...) –  RecklessReckoner Mar 15 '14 at 4:43
$\displaystyle\sum_{k=1}^nk^p=\mathcal O\Big(n^{p+1}\Big)$. See Faulhaber's formulas. –  Lucian Mar 15 '14 at 4:51

Consider $$\frac{1}{n}\sum_1^n \left(\frac{k}{n}\right)^{15}.\tag{1}$$ This is a right Riemann sum for $$\int_0^1 x^{15}\,dx.$$ The limit as $n\to\infty$ of (1) exists. From that, you should be able to find the answers to both questions.
Remark: We used a Riemann sum, since that seemed to be the approach requested. Another way of viewing things is that $\sum_1^n k^{15}$ is a polynomial in $n$ of degree $16$. Thus the first sequence obviously diverges to $\infty$, and the second converges to $0$.
Im confused with the power of k and n since I cannot do what you did here, just take 1/n out. That doesnt work. –  Danny Mar 15 '14 at 6:11
I take it you agree that the limit for the expression I wrote down, the one with altogether an $\frac{1}{n^{16}}$, exists and is non-zero. It follows that if you divide by one fewer $n$, as in the first problem, the limit is $\infty$, or if you prefer, doesn't exist. And if you divide by one more $n$, as in the second problem, the limit is $0$. Clear? –  André Nicolas Mar 15 '14 at 6:23