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.

  • $\begingroup$ 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...) $\endgroup$ – colormegone Mar 15 '14 at 4:43
  • $\begingroup$ $\displaystyle\sum_{k=1}^nk^p=\mathcal O\Big(n^{p+1}\Big)$. See Faulhaber's formulas. $\endgroup$ – 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 $0$.

  • $\begingroup$ This is an easier case where we can clearly write the function, but what I have above is different than this. please take another look on the the question above especially the power of n and k. $\endgroup$ – Danny Mar 15 '14 at 6:09
  • $\begingroup$ 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. $\endgroup$ – Danny Mar 15 '14 at 6:11
  • $\begingroup$ 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? $\endgroup$ – André Nicolas Mar 15 '14 at 6:23
  • $\begingroup$ yes, thanks a lot :) $\endgroup$ – Danny Mar 15 '14 at 16:48
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Mar 15 '14 at 16:58

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