malalawak44

2022-07-13

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

Kiana Cantu

To reason here, imagine expanding the sum out. You would have
$\sum _{n\in \mathbb{N}}\left(\mathrm{log}\sqrt{n+1}-\mathrm{log}\sqrt{n}\right)=\left(\mathrm{log}\sqrt{2}-0\right)+\left(\mathrm{log}\sqrt{3}-\mathrm{log}\sqrt{2}\right)+\left(\mathrm{log}\sqrt{4}-\mathrm{log}\sqrt{3}\right)+\dots$
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:
${S}_{N}=\sum _{n=1}^{N}\left(\mathrm{log}\sqrt{n+1}-\mathrm{log}\sqrt{n}\right)=\mathrm{log}\sqrt{2}+\cdots +\left(\mathrm{log}\sqrt{n}-\mathrm{log}\sqrt{N-1}\right)\phantom{\rule{0ex}{0ex}}+\left(\mathrm{log}\sqrt{N+1}-\mathrm{log}\sqrt{N}\right)=\mathrm{log}\sqrt{N+1}$
Taking the limit, we can see
$\underset{N\to \mathrm{\infty }}{lim}{S}_{N}=\sum _{n\in \mathbb{N}}\left(\mathrm{log}\sqrt{n+1}-\mathrm{log}\sqrt{n}\right)=\underset{N\to \mathrm{\infty }}{lim}\mathrm{log}\sqrt{N+1}=\mathrm{\infty }$
So, this series will diverge with its partial sums.

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