# Evaluate $\lim\limits_{n \to \infty} \frac 1{n\log \left ( 1+\frac1n \right ) }$

$$\lim\limits_{n \to \infty} \frac 1{n\log \left ( 1+\frac1n \right ) }$$

I start by writing $$\lim\limits_{n \to \infty} \frac 1n=0$$, and $$\lim\limits_{n \to \infty} \frac 1{n+\log \left ( 1\frac1n \right ) }=0$$

Since limit exists, by multiplication law, $$\lim\limits_{n \to \infty} \frac 1{n\log \left ( 1+\frac1n \right ) }=0$$

Hence the series $$\sum_{n}^{\infty }\frac{1}{n\log 1+\frac{1}{n}}$$ converge.

However, the $$\lim n \to \infty \frac{1}{n\log \left ( 1+\frac{1}{n} \right ) }$$ is given to be 1 in the solution, would really like to know where I've done wrong.

Thanks for the help.

• The second limit you wrote is $\infty$ I believe.
– dfnu
Dec 7, 2019 at 8:57
• You have two serious reasoning flaws: the first at line 3 "Since ..." . That's a non sequitur . The second and probably most serious one is at line 4: you seem to believe that if $\;\lim a_n =0\;$ then $\;\sum\limits_{n=1}^\infty a_n\;$ converges. This is completely false. Dec 7, 2019 at 9:29

You are right: $$\lim\limits_{n \to \infty} \frac 1{n+\log \left ( 1+\frac1n \right ) }$$ is $$0$$, since we end up with $$\frac{1}{\infty}$$

If we instead consider the $$\lim\limits_{n \to \infty} \frac 1{n\log \left ( 1+\frac1n \right ) }$$, well this converges to 1: in fact, the denominator could be rewritten by the properties of logarithm as $${\ln \left ( 1+\frac1n \right )^n },$$ which is, as $$n \to\ \infty$$, $$ln( e)$$, that is $$1$$. So, eventually $$\frac {1}{1} = 1$$, and the limit is done!

• To say "youre right" seems to imply that the poster is right in all she wrote...which is not. Read comments. Dec 7, 2019 at 9:31
• true, thanks now I'll edit Dec 7, 2019 at 9:35

Your first limit $$\lim_{n\to\infty}\frac{1}{n+\log(1+\frac{1}{n})}=0$$ In the corrected case we have $$\lim_{n\to \infty}\frac{1}{\log\left(1+\frac{1}{n}\right)^n}=1$$ since $$\lim_{n\to \infty}\log\left(1+\frac{1}{n}\right)^n=\log(e)=1$$ if $$\log$$ means logarithmus naturalis, to the base $$e$$

• Oh my mistake... It's suppose to be multiplying
– user596652
Dec 7, 2019 at 9:14

We have that

$$n+\log \left ( 1+\frac1n \right )=n+\frac1n\frac{\log \left ( 1+\frac1n \right )}{\frac1n} \to \infty+0\cdot 1=\infty$$

and therefore

$$\lim\limits_{n \to \infty} \frac 1{n+\log \left ( 1+\frac1n \right ) }=0$$

which is indeed correct.

For the other one we have

$$\lim\limits_{n \to \infty} \frac 1{n\log \left ( 1+\frac1n \right ) }=\lim\limits_{n \to \infty} \frac 1{\log \left ( 1+\frac1n \right )^n }=\frac1{\log e}=1$$

and therefore the series $$\sum_{n}^{\infty }\frac{1}{n\log 1+\frac{1}{n}}$$ diverges since the necessary condition $$a_n \to 0$$ is not satisfied.