# Convergence of $\sum_{n=1}^\infty \frac{1}{n}-\ln(1+ \frac{1}{n})$

I'm having trouble to prove/disprove the Series $$\sum_{n=1}^\infty\left( \frac{1}{n}-\ln(1+ \frac{1}{n})\right)$$ being convergent or divergent.

I tried the ratio test but it seems to be inconclusive. I think I should use here some reference for the comparison test but I couldn't find any matching Series.

Also tried to make a common denominator and received $\sum_{n=1}^\infty \frac{1-n\ln(1+ \frac{1}{n})}{n}$ so I have to find something smaller then $n\ln(1+ \frac{1}{n})$, and I couldn't figure out the matching alternative.

• $\frac1n - \log\left(1 +\frac1n\right) = \frac1n - \int_n^{n+1}\frac{dx}{x} \le \frac1n - \frac1{n+1}$ – achille hui Mar 6 '16 at 13:38
• @achillehui Please make that an answer! – RKD Mar 6 '16 at 13:40
• @achillehui how does that come to be that $\frac{1}{(n+1)}<ln(1+ \frac{1}{n})$. I know that $ln(1+X)<X$ always. thus $\frac{1}{(n)}>ln(1+ \frac{1}{n})$ and $\frac{1}{n+1}$ is very close to $\frac{1}{n}$ for high n.. – Ami Gold Mar 6 '16 at 13:42
• @Ami $\frac{1}{x}$ is a strictly decreasing function, so $\int_n^{n+1} \frac{dx}{x} > \int_n^{n+1}\frac{dx}{n+1} = \frac{1}{n+1}$. – achille hui Mar 6 '16 at 13:44

Note that $\ln (1+x)=x-\frac12x^2+O(x^3)$. Hence $\frac1n-\ln(1+\frac 1n)=\frac1{2n^2}+O(n^{-3})$ and so for $n$ large enough we have $0<\frac1n-\ln(1+\frac 1n)<\frac1{n^2}$.

In case you are "afraid" of the big-O or did not learn about Taylor expansion yet, it is sufficient to show the following: $$x-x^2\le \ln(1+x)\le x\qquad\text{for }x\ge0.$$ The right hand side is an immediate consequence of $e^x\ge 1+x$ (the mother of all inequalities for the exponential). The left hand side follows from $e^{x^2-x}\ge 1+x^2-x>0$ so that $$e^{x-x^2}=\frac1{e^{x^2-x}}\le \frac1{1-x+x^2} =\frac{1+x}{1+x^3}\le 1+x$$

• Ok, so how it comes that $e^{x^2-x}\ge 1+x^2-x>0$ leads to $$x-x^2\le \ln(1+x)\le x\qquad\text{for }x\ge0.$$ As I understand, it leads to that $ln(1+(x-x^2))\le x-x^2$ – Ami Gold Mar 6 '16 at 13:59

We'll use the integral test:

$$\sum_{n=1}^{\infty} \left( \frac{1}{n} - \ln\left(1+\frac{1}{n}\right) \right)$$

converges if and only if the integral converges. $$\int_1^\infty \frac{1}{x}-\ln\left(1+\frac{1}{x}\right)dx = \lim_{x\to\infty} (x+1)\ln\left(\frac{x}{x+1}\right) - 2\ln\left(\frac{1}{2}\right) = 2\ln(2)-1 \neq \infty$$ So it converges.

Ok, after being bloody stupid I figured out what was my mistake.

So at first I tried to use the limit comparison test with $\frac{1}{n^2}$ but somehow I calculated the limit wrongly, thus I got stuck.

but now after I calculated once again the limit it showed me that I was right, take a look:

$$lim_{n \to \infty} \frac{\frac{1}{n}-ln(1+\frac{1}{n})}{\frac{1}{n^2}}=|L'Hopitall|=lim_{n \to \infty} \frac{\frac{-1}{n^3+n^2}}{\frac{-2}{n^3}}= \frac{1}{2}lim_{n \to \infty} \frac{n^3}{n^3+n} \to \frac{1}{2}$$

Then by the limit comparison test, we get that both Series converge or diverge together. but because $\frac{1}{n^2}$ converge, then also the original Series Converge(!). Q.E.D

$*$where is the face-palm smiley...$*$

The natural logarithm function is unique in that it is guaranteed that at a certain value $x = c,$ it will always grow slower than a power function (such as $f(x) = x^{2}$) for $x \gt c.$ So it can be seen that eventually, $$0 \lt \frac{1}{n} - \ln\left(1 + \frac{1}{n}\right) \lt \frac{1}{n^{p}},$$ where $p$ is a number larger than $1.$ By the p-series test in combination with the direct comparison test, the series converges.

• I can agree with that. but is there a more "formal" way to show that $\frac{1}{n} - \ln(1 + \frac{1}{n}) < \frac{1}{n^{p}}$ (when p>1) – Ami Gold Mar 6 '16 at 13:46

Just use the limit comparison test . Use the series $\sum_{n=1}^\infty(1/n)$. The limit comes out to be 1 and hence both the series converge or diverge together.

• no, the limit would be 0. and thus it would be inconculsive.. I tried that – Ami Gold Mar 6 '16 at 13:36
• Pretty sure that limit comes out to $0$. Edit: Whoops, OP got there before me. – RKD Mar 6 '16 at 13:36
• Ohh yes i am sorry.then the other answer seems right.. – Upstart Mar 6 '16 at 13:37