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I would like to know if my reasoning for the following questions is correct.

In each case determine whether or not the series converges

$$ 1. \quad \underset{k=1}{\overset{\infty}{\sum}} (-1)^k \cos\left(\frac{1}{k}\right)$$ $$ 2. \quad \underset{k=1}{\overset{\infty}{\sum}} \frac{\sin \left(\frac{k \pi}{2}\right)}{\sqrt{k}}$$ $$ 3. \quad \underset{k=1}{\overset{\infty}{\sum}} \frac{1}{k^{\left(1 + 1/k\right)}}$$

I have reasoned the following way:

  1. We have the (contrapositive statement) that if a series terms do not approach zero, then the series does not converge. We know that $\underset{k \rightarrow \infty}{\lim} (-1)^k \cos\left(\frac{1}{k}\right) \rightarrow \pm \cos(0) = \pm 1$. Hence the series diverges.

  2. Here we use Weierstrass's M-test. $$\left| \frac{\sin \left(\frac{k \pi}{2}\right)}{\sqrt{k}} \right| \leq \frac{1}{\sqrt{k}}$$ Since $\underset{k=1}{\overset{\infty}{\sum}} \frac{1}{\sqrt{k}} < \infty$, then $\underset{k=1}{\overset{\infty}{\sum}} \frac{\sin \left(\frac{k \pi}{2}\right)}{\sqrt{k}}$ converges.

  3. This is the one I'm really struggling with. Intuitively, it feels like the series should diverge, since as $n \rightarrow \infty$, the series looks like the harmonic series. But using the ratio test (which should be possible since the series is positive) we get: $$\frac{k^{\left(1 + \frac{1}{k}\right)}}{(k+1)^{\left(1 + \frac{1}{k+1}\right)}} < 1$$ So the series should converge?

What I would like to know is:

  1. Is my reasoning for these questions correct?
  2. In the case when intuition and convergence tests contradict each other, how can we be certain that we have arrived at the correct answer?

Thanks!

EDIT: So $\sum \frac{1}{\sqrt{k}} = \infty$, so my reasoning for Q2 is obviously not correct.

user9750060
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    $\sum \frac{1}{\sqrt{k}}$ is divergent... so your reasoning does not work here. But you can use criteria for alternating series. $\sin \frac{k\pi}{2} = \cdots$ – PierreCarre Oct 13 '20 at 13:35
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    If convergence tests contradict each other, it is user error. This ain't physics :) Note: this is actually useful information because it alerts you to the fact that you have made an error, as we all sometimes do. – Thomas Winckelman Oct 13 '20 at 13:36

1 Answers1

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For (1), the sequence does not approach zero because for $k \geq 1,$ $\cos(1/k) \geq \cos(1)$, hence $\lvert a_k - 0\rvert = \lvert a_k\rvert \geq \cos(1) > 0$. It is also true that $\lvert a_k\rvert\to 1$, as you noticed, which also prevents the terms from approaching zero.

For (2), you have what is essentially an alternating sequence, except that instead of $+,-,+,-,\ldots$ it is $+,0,-,0,+,0,-,\ldots$.

For (3), it is true that for any positive integer $k$, $$\frac{k^{1 + 1/k}}{(k+1)^{1 + 1/(k+1)}} < 1,$$ but this is not the ratio test; the ratio test is about the limit of the ratio. And $$\lim_{k\to\infty} \frac{k^{1 + 1/k}}{(k+1)^{1 + 1/(k+1)}} = 1,$$ so the ratio test is inconclusive. You might try a comparison test instead. You may have noticed that $k^{1 + 1/k}$ is larger than $k$. But how much larger does it get?

David K
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