Test the convergence of the series $\sum_{n=1}^{\infty}a_n$ , where

$$a_n=\begin{cases}\dfrac{1}{n^2} & \text{ if $n$ is not a square integer},\\[6pt] \dfrac{1}{n^{2/3}} & \text{ if $n$ is a square integer}.\end{cases}$$

I can not understand that how I approach. I think it may be by Integral Test. But I am unable to apply this test.

  • $\begingroup$ Integral test does not apply because we require the underlying function, hence the sequence, to be monotone. $\endgroup$ – Theo Bendit Jul 20 '15 at 16:48
  • $\begingroup$ Split into two subseries. If both converge, so does their sum (it is positive terms, so convergence is absolute convergence, and thus reordering terms is allowed). $\endgroup$ – vonbrand Jul 23 '15 at 15:58

This is a series whose terms are all positive which means it converges if and only if it absolutely converges. Therefore we can break it up into two series and test their convergence. $$\sum_{n=1}^\infty a_n\leq\sum_{n=1}^\infty\frac{1}{n^2}+\sum_{n=1}^\infty \frac{1}{n^{4/3}}=\frac{\pi^2}{6}+\zeta(4/3)<\infty.$$

  • $\begingroup$ When $n$ is square integer then $a_n=\frac{1}{n^{2/3}}$...NOT $4/3$ $\endgroup$ – Empty Jul 20 '15 at 16:59
  • 1
    $\begingroup$ But we are summing over all numbers of the form $n=k^2$, so the sum is $$\sum_{\text{n is square}}\frac{1}{n^{2/3}}=\sum_{k=1}^\infty\frac{1}{(k^2)^{2/3}}= \sum_{k=1}^\infty\frac{1}{k^{4/3}}.$$ $\endgroup$ – Alex S Jul 20 '15 at 17:03

I'm not sure which test I'd use per se, but the way I'd establish it is as follows: $$\sum_{k=1}^n a_n \le \sum_{k=1}^n k^{-2} + \sum_{1 \le k \le \sqrt{n}} k^{-4/3} \le \sum_{k=1}^\infty k^{-2} + \sum_{k = 1}^\infty k^{-4/3} < \infty,$$ hence the partial sums are a montone increasing sequence that's bounded above, and hence convergent.


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