I have to examine convergence of series $\sum_{n=1}^{\infty}{\frac{\ln(n+1)-\ln(n)}{\sqrt{n}}}$ but I even don't know how to begin. Any hints?
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By the MVT, we have $\frac{\ln(n+1)-\ln n}{1}=\frac{1}{x_n}$ where $n\lt x_n\lt n+1$. – André Nicolas Mar 22 '15 at 20:19
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Hint: $\log:(0,\infty)\to \Bbb R$ is differentiable, take $x=n+1$ and $y=n$, then by the Mean Value Theorem, there is a $c\in (n+1,n)$ such that $\log(n+1)-\log(n)=\frac{1}{c}(n+1-n)=\frac{1}{c}<\frac{1}{n}$. Hence $\dfrac{\log(n+1)-\log(n)}{\sqrt n}<\dfrac{1}{n^{3/2}}$, for each $n\in \Bbb N$. So the series converges (the given series is dominated by the convergent series $\sum\limits_{n=1}^\infty{\dfrac{1}{n^p}}$, where $p>1$).
user149418
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Just see this, for large $n$ we have (using Taylor series of $\ln(1+t)$)
$$ \frac{\ln(n+1)-\ln(n)}{\sqrt{n}} = \frac{\ln(1+1/n)}{\sqrt{n}}\sim \frac{1/n}{\sqrt{n}} = \frac{1}{n^{3/2}}.$$
science
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