Showing that $\ln(1+(1/n))$ diverges with terms approaching 0

Question

For the harmonic series where $$\sum_{n=1}^{\infty}(1/n)$$ diverges whose terms approach zero as '$n$' goes it infinity, I was meant to show that $$\sum_{n=1}^{\infty}\ln(1+(1/n))$$ also has the same property.

I first started with simplifying the series to $$\sum_{n=1}^{\infty}\ln((1+n)/n) \ ,$$which as we know is the same as $$\sum_{n=1}^{\infty}\ln(1+n)-\ln(n)$$ as $$\ln(a/b) = \ln(a)-\ln(b)\ .$$

I then used the telescoping series formula to simplify $$\sum_{n=1}^{\infty}\ln(1+n)-\ln(n)$$ to $$\sum_{n=1}^{\infty}-\ln(1)+\ln(N+1)\ .$$

This then simplifies to $$\sum_{n=1}^{\infty}0+\ln(N+1)\ .$$

After we reach this point, wouldn't we able to show that if we let '$n$' go to infinity, we see that the limit of $$\ln(N+1)$$ as '$n$' goes to infinity, we see that the series diverges and the terms approach $0$? Is this a proper way to show that this in fact has the same properties as the harmonic series (divergence series with terms approaching $0$). Is this way of showing the property valid? Thanks.

Answer 1

Since$$\sum_{n=1}^\infty\log(n+1)-\log(n)$$is a telescoping sries, you now use the fact that$$(\forall N\in\mathbb N):\sum_{n=1}^N\log(n+1)-\log(n)=\log(N+1)-\log(1)=\log(N+1).$$Since $\lim_{N\to\infty}\log(N+1)=\infty$, your series diverges.

Answer 2

Consider $f(x)=\log(1+x)-x/2$; then $$ f'(x)=\frac{1}{1+x}-\frac{1}{2}=\frac{1-x}{1+x} $$ which is positive for $-1<x<1$. Since $f(0)=0$, we conclude that $f(x)>0$ for $0<x<1$ and therefore $$ \log\biggl(1+\frac{1}{n}\biggr)>\frac{1}{2n} $$

In a (not so) different way: $$ \lim_{n\to\infty}\frac{\log(1+1/n)}{1/n}=\lim_{x\to0}\frac{\log(1+x)}{x}=1 $$ so the two series have the same character. This also explain why it is possible to find a constant $k$ such that $$ k\frac{1}{n}<\log\biggl(1+\frac{1}{n}\biggr) $$