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I know the series converges. I am not allowed to use any calculus to solve this problem. My professor suggested that I try and use the limit comparison test, but I do not know how to evaluate any limits involving these terms.

Jungleshrimp
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  • Multiplying with $$\frac{\frac{1}{\sqrt{n}} + \sqrt{\ln (1 + \frac{1}{n})}}{\frac{1}{\sqrt{n}} + \sqrt{\ln (1 + \frac{1}{n})}}$$ may help. – Daniel Fischer Nov 13 '19 at 20:58
  • What is the definition of $\log$ that is used in the course? Looks like we might need to get to the second order term of $\log(1+x)$ expansion (or equivalent approaches like using $\log(1+x) \ge x - \frac{x^2}{2}$) to eliminate the $\frac{1}{\sqrt{n}}$ term. – Aryabhata Nov 14 '19 at 02:27
  • In rudin, it first defines e as a limit of (1+1/n)^n, then says "log(n)" is the logarithm of n to the base e. – Jungleshrimp Nov 14 '19 at 04:21

2 Answers2

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What you could do is to use the Taylor expansion $$\log \left(1+\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ Now, using the binomial expansion $$\sqrt{\log \left(1+\frac{1}{n}\right)}=\frac 1 {\sqrt n}-\frac 14 \frac 1 {n^{3/2}}+O\left(\frac{1}{n^{5/2}}\right)$$ $$\frac 1 {\sqrt n}-\sqrt{\log \left(1+\frac{1}{n}\right)}=\frac 14 \frac 1 {n^{3/2}}+O\left(\frac{1}{n^{5/2}}\right)$$

Claude Leibovici
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$\displaystyle 0<\sum\limits_{n=1}^\infty \left(\frac{1}{\sqrt{n}}-\sqrt{\ln\left(1+\frac{1}{n}\right)}\right)=\sum\limits_{n=1}^\infty\frac{1-\ln\left(1+\frac{1}{n}\right)^n}{\sqrt{n}\left(1+\ln\left(1+\frac{1}{n}\right)^n\right)}<$

$\displaystyle <\,\sum\limits_{n=1}^\infty\frac{\ln\left(1+\frac{1}{n}\right)^{n+1}-\ln\left(1+\frac{1}{n}\right)^n}{\sqrt{n}}=\sum\limits_{n=1}^\infty\frac{\ln\left(1+\frac{1}{n}\right)}{\sqrt{n}}\leq\sum\limits_{n=1}^\infty\frac{1}{n\sqrt{n}}=\zeta(\frac{3}{2})$

Used:

$\displaystyle \frac{1}{1+\ln\left(1+\frac{1}{n}\right)^n}<1~~$ and $~~\displaystyle 1\leq\ln\left(1+\frac{1}{n}\right)^{n+1}~~$ and $~~\displaystyle \ln\left(1+\frac{1}{n}\right)\leq\frac{1}{n}~$ for all $~n\in\mathbb{N}$

user90369
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