Methods for showing divergence of $\sum_{1}^{\infty} ln(1+\frac{1}{n})$?

[ognik]

Does\: $ \sum_{1}^{\infty} ln(1+\frac{1}{n}) $\: converge?
I tried the limit comparison test with b_n=1/n and got that it diverges, which also looks right.
However I also tried the ratio test:
$ \lim_{{n}\to{\infty}} \left| \frac{{a}_{n+1}}{{a}_{n}} \right| = \lim_{{n}\to{\infty}} \left| \frac{\ln\left({1+\frac{1}{n+1}}\right)}{\ln\left({1+\frac{1}{n}}\right)} \right| = ? $
I have edited my original post because I had, thoughtlessly, used Ln a/Ln b = Ln (a-b), which is a silly mistake, of course that's not true.
Actually one should use L'Hositals rule to simplify this, which I am currently busy with :-).

 

[ognik]

Using L'H rule I got that limit = 1, which is inconclusive for the ratio test, so the limit comparison test seems the only way to do this one.

 

[GJA]

Hi ognik,

so the limit comparison test seems the only way to do this one.

Limit comparison test certainly works. Thought I might offer two other methods that do as well:

1) Write $\ln\left(1+\frac{1}{n}\right)=\ln\left(\frac{n+1}{n}\right)=\ln(n+1)-\ln(n),$ then use a sequence of telescoping partial sums to show divergence.

2) It's useful to note that
$$\ln(x)=\int_{1}^{x}\frac{1}{t}dt$$
From this it follows that
$$\ln\left(1+\frac{1}{n}\right)=\int_{1}^{1+\frac{1}{n}}\frac{1}{t}dt\geq \frac{1}{n+1},$$
where the inequality holds from the shape of the graph of $f(t)=\frac{1}{t}.$

Hopefully this has been helpful. If not, feel free to disregard this. Let me know if anything is unclear/not quite right.

 

[ognik]

Hi ognik,
Limit comparison test certainly works. Thought I might offer two other methods that do as well:

1) Write $\ln\left(1+\frac{1}{n}\right)=\ln\left(\frac{n+1}{n}\right)=\ln(n+1)-\ln(n),$ then use a sequence of telescoping partial sums to show divergence.

2) It's useful to note that
$$\ln(x)=\int_{1}^{x}\frac{1}{t}dt$$
From this it follows that
$$\ln\left(1+\frac{1}{n}\right)=\int_{1}^{1+\frac{1}{n}}\frac{1}{t}dt\geq \frac{1}{n+1},$$
where the inequality holds from the shape of the graph of $f(t)=\frac{1}{t}.$

Hopefully this has been helpful. If not, feel free to disregard this. Let me know if anything is unclear/not quite right.

Both are useful and appreciated, thanks.