I'm learning calculus, specifically convergence of series, and need with with the following exercise:

Examine convergence and absolute convergence of the given series:

$$(a) \sum_{n=1}^{\infty}\frac{n\sqrt[n]{n}}{n^2+1} \quad \quad (b) \sum_{n=2}^{\infty}\frac{\cos(n)}{n\log^3(n)}.$$

Note: I'm not allowed to use the integral test.

Here's my work so far: The Root and the Ratio tests leads to the indeterminate case where the limit is equal to $1$ for series $(a)$ so my next thought was to use the Comparison test. For $(a)$ one has

$$\sum_{n=1}^{\infty}\left|\frac{n\sqrt[n]{n}}{n^2+1}\right| = \sum_{n=1}^{\infty}\frac{n\sqrt[n]{n}}{n^2+1}$$


$$\frac{n\sqrt[n]{n}}{n^2+1} \leq \frac{n\sqrt[n]{n}}{n^2} = \frac{\sqrt[n]{n}}{n} \implies \sum_{n=1}^{\infty}\frac{n\sqrt[n]{n}}{n^2+1} \leq \sum_{n=1}^{\infty}\frac{\sqrt[n]{n}}{n}.$$

For $(b)$ we have

$$\sum_{n=2}^{\infty}\left|\frac{\cos(n)}{n\log^3(n)}\right| \leq \sum_{n=2}^{\infty}\frac{1}{n\log^3(n)}.$$

Is my work correct so far? How do I examine the convergence of the RHS series that I found?

  • $\begingroup$ @RobertZ Yes, absolutely. I corrected my original post. Thank you. $\endgroup$ – Von Kar Sep 12 '16 at 9:38
  • $\begingroup$ Ok. Does my answer clear your doubts? $\endgroup$ – Robert Z Sep 12 '16 at 9:43
  • $\begingroup$ @RobertZ Yes, your answer was extremely helpful. I posted my answer for $(b)$ using the Cauchy Condensation Theorem as you proposed. Thanks again. $\endgroup$ – Von Kar Sep 12 '16 at 12:13

As regards (a), since $\sqrt[n]{n}=e^{\frac{\ln n}{n}}\to 1$, then $$\frac{n\sqrt[n]{n}}{n^2+1}\sim \frac{n\cdot 1}{n^2}=\frac{1}{n}$$ so your series is not convergent by the Asymptotic Comparison Test and the fact that $\sum_{n\geq 1}\frac{1}{n}=+\infty$.

As regards (b), as you noted $$\sum_{n=2}^{\infty}\left|\frac{\cos(n)}{n\log^3(n)}\right| \leq \sum_{n=2}^{\infty}\frac{1}{n\log^3(n)}.$$ By the Comparison Test, this implies that your series is convergent because $$\sum_{n=2}^{\infty}\frac{1}{n\log^a(n)}$$ is convergent for all $a>1$. In order to prove it consider the integral $$\int_{x=2}^{\infty}\frac{dx}{x\log^a(x)}=\left[\frac{\log^{1-a}(x)}{1-a}\right]_2^{+\infty}<+\infty,$$ or apply the Cauchy Condensation Test.


Using the Cauchy Condensation Theorem for $(b)$ as suggested by Robert Z, one has

$$\sum_{n=2}^{\infty} \frac{1}{n(\log n)^3} \sim \sum_{n=2}^{\infty} 2^n \frac{1}{2^n(\log 2^n)^3} = \sum_{n=2}^{\infty} \frac{1}{(n \log 2)^3} = \left(\frac{1}{\log 2}\right)^3 \sum_{n=2}^{\infty} \frac{1}{n^3} < +\infty$$

which converges by the p-test. Therefore, the original series is absolutely convergent.

  • 1
    $\begingroup$ That is correct. Well done!! $\endgroup$ – Robert Z Sep 12 '16 at 12:15

for a) i would write $$\frac{n\cdot n^{1/n}}{n^2+1}\geq \frac{1}{n+1}$$ this is equivalent ot $$n\geq 1$$ which is true.


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