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This was a practice question.

Does this infinite sum converge or diverge? $$\sum_{n=1}^{\infty}(\ln {\sqrt {n+1}} - \ln{\sqrt{n}})$$

I tried applying Divergence test, but since the limit equals 0, it's inconclusive.

Then I tried telescoping series, but quickly realized it goes to infinity, so that wasn't helpful either.

So does it converge or diverge? If it converges, what does it converge to?

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    $\begingroup$ Do you know about telescoping series? Can you tell what $\sum\limits_{n = 1}^N {(\ln \sqrt {n + 1} - \ln \sqrt n )}$ is? $\endgroup$
    – Gary
    Mar 21, 2022 at 4:33
  • $\begingroup$ @Gary Yeah, I got what beeselmane got, but just gave up after realizing it will never end. But it makes sense. Thanks $\endgroup$
    – user842157
    Mar 21, 2022 at 4:44

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To reason here, imagine expanding the sum out. You would have

$$ \sum_{n \in \mathbb{N}} (\log \sqrt{n+1} - \log\sqrt{n}) = (\log \sqrt{2} - 0) + (\log \sqrt{3} - \log \sqrt{2}) + (\log \sqrt{4} - \log \sqrt{3}) + \dots $$

As you can see, if you look at each pair of terms, the term added by the previous term will be taken away by the subtraction on the next term. You can then see the convergence by simply taking the limit of the partial sums:

$$ S_{N} = \sum_{n = 1}^{N}(\log \sqrt{n+1} - \log\sqrt{n}) = \log \sqrt{2} + \dots + (\log \sqrt{n} - \log \sqrt{N - 1}) \\+ (\log \sqrt{N + 1} - \log \sqrt{N}) = \log\sqrt{N + 1} $$

Taking the limit, we can see

$$ \lim_{N \to \infty} S_{N} = \sum_{n \in \mathbb{N}} (\log \sqrt{n+1} - \log\sqrt{n}) = \lim_{N \to \infty}\log\sqrt{N + 1} = \infty $$

So, this series will diverge with its partial sums.

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  • $\begingroup$ Use \log to produce $\log$. It looks much better. $\endgroup$
    – Gary
    Mar 21, 2022 at 4:45
  • $\begingroup$ Nice way to bypass $\sum 1/n$ diverges. $\endgroup$ Mar 21, 2022 at 4:57