# Prove $\sum\limits_{n=1}^\infty \ln({1+\frac 1 {{n}}})$ is divergent. [duplicate]

Evaluate if the following series is convergent or divergent: $\sum\limits_{n=1}^\infty \ln({1+\frac 1 {{n}}})$.

I tried to evaluate the divergence, applying the Weierstrass comparison test:

$\sum\limits_{n=1}^\infty \ln({1+\frac 1 {{n}}})>\sum\limits_{n=1}^\infty \ln({\frac 1 {{n}}})$. Since the function is decreasing then as $1+\frac{1}{n}$ is closer to $0$ then the inequality follows.

Obviously $\sum\limits_{n=1}^\infty \ln({\frac 1 {{n}}})$ diverges since $\lim_{n\to\infty}\ln({\frac 1 {{n}}})=-\infty$.

Questions:

1) Is my answer right? If not why?

2) What other kind of approach do you propose?

## marked as duplicate by Xander Henderson, Namaste calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 31 '18 at 17:11

• – projectilemotion Mar 31 '18 at 16:39
• Showing that something is greater than $-\infty$ doesn't say much by itself. – alex.jordan Mar 31 '18 at 16:41

$$\sum_{n=1}^N\ln\left(1+\frac{1}{n}\right)=\ln(N+1)\to\infty\textrm{ as }N\to\infty$$

• How do I prove $\sum_{n=1}^N\ln\left(1+\frac{1}{n}\right)=\ln(N+1)$? – Pedro Gomes Mar 31 '18 at 17:05
• $\sum_{n=1}^N\ln\left(1+\frac{1}{n}\right)=\ln(\frac{2}{1})+\ln(\frac{3}{2})+\ln(\frac{4}{3})+\cdots+\ln(\frac{N+1}{N})=\ln(\frac{2}{1}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdots\frac{N+1}{N})=\ln(N+1)$ – CY Aries Mar 31 '18 at 17:08

Let S be the sum of the series. The comparison you used only tells you that $S > -\infty,$ which is useless.

To determine convergence, try using the fact that $\ln a + \ln b = \ln(ab).$ Notice that $1+1/n = \frac{n+1}{n}.$

• I end up with $\sum_\limits{n=1}^{\infty}(\ln(n+1)-\ln(n))$.How can I prove with this expression that series diverge? Thanks for your answer! – Pedro Gomes Mar 31 '18 at 17:36

1) Your answer is not correct. The only way you can say that $\sum_{n=1}^{\infty} a_n$ diverges because $\sum_{n=1}^{\infty} b_n$ diverges and $a_n\geq b_n$ (comparison test) is if $\sum_{n=1}^{\infty} b_n \to \infty$. For a concrete example, take $a_n=0$ and $b_n=-1$.

2) I would recommend using the limit comparison test on $\ln(1+1/n)$ and $1/n$.

Isn't right because $\log(1/n)<0$, But you can use that $$n\to\infty\implies \log(1+1/n)\approx\frac1n.$$

1) Well, you haven't shown that $\sum\limits_{n=1}^\infty \ln({1+\frac 1 {{n}}})$ diverges. You've show that the sum is $\ge -\infty$ but that is true of all series, right?

2) Try comparing $\ln(1+\frac{1}{n})$ to $\frac{1}{n}.$

HINT: $$\lim_{n\to\infty}\frac{\ln(1+(1/n))}{\frac1n}=1.$$

And your evaluation of strictly positive series by negative series from below is incorrect (useless).

use the inequality $\text{log}(1+x) > \frac{x}{1+x}$ for any nonzero $x \geq -1$ and you get $\text{log}(1+1/n)> \frac{1}{1+ n}$. Summing over both sides over all $n \in \mathbb{N}$ yields the harmonic series on the right, which does not converge.

Hint: Assume that $p_n\ge0$ with $p_n\to 0$ as $n\to\infty$. Then, by $$\lim_{n\to \infty}\frac{\log(1+p_n)}{p_n}=\lim_{x\to 0}\frac{\log(1+x)}{x}=1$$ and the limit comparison test you have that $\sum_{n=1}^{\infty}\log(1+p_n)$ and $\sum_{n=1}^{\infty}p_n$ are equivalent, meaning that either both converge or both diverge.