Q:

Prove the convergence of series $\sum_{k=1}^{\infty}\log(1+\frac{1}{\sqrt{k}})$ by Cauchy criterion

Given $$\sum_{k=1}^{\infty}\log\left(1+\frac{1}{\sqrt{k}}\right)$$
and by definition I need to prove that for $\forall \epsilon>0, \exists n_0 \text{ s.t. } \forall n>n_0, \forall p=1,2,...$
\begin{align}
\left|\sum_{k=n+1}^{n+p}\log(1+\frac{1}{\sqrt{k}}) \right |<\epsilon\end{align}
I expanded the sum but wasn't able to get final result for right $n_0$ depending on $\epsilon$. Here's how far I came
\begin{align}\sum_{k=n+1}^{n+p}\log\frac{\sqrt{k}+1}{\sqrt{k}}=\log\frac{(\sqrt{n+1}+1)(\sqrt{n+2}+1)...(\sqrt{n+p}+1)}{\sqrt{(n+1)(n+2)...(n+p)}}\end{align}
taking care of $p$ by choosing it $p=n$ we're having (I hope I haven't done a mistake)
\begin{align}\leq\log\frac{(\sqrt{2n}+1)}{\sqrt{2n}}<\epsilon\end{align}
Any ideas how to continue after this?

A:

Your series is not convergent. $\log(1+x)$ is a concave function on the interval $[0,1]$, hence:
$$\forall x\in[0,1],\qquad \log(1+x)\geq x\log 2\tag{1} $$
holds. That gives:
$$\begin{eqnarray*}\sum_{n=1}^{N}\log\left(1+\frac{1}{\sqrt{n}}\right)&\geq& \log 2\cdot\sum_{n=1}^{N}\frac{1}{\sqrt{n}}\\&\geq& 2\log 2\cdot\sum_{n=1}^{N}\left(\sqrt{n+1}-\sqrt{n}\right)=\color{red}{\log 4\cdot\left(\sqrt{N+1}-1\right)}\tag{2}\end{eqnarray*} $$
that grows unbounded as $N\to +\infty$.

