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

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.

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    $\begingroup$ The second limit you wrote is $\infty$ I believe. $\endgroup$
    – dfnu
    Dec 7, 2019 at 8:57
  • $\begingroup$ 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. $\endgroup$
    – DonAntonio
    Dec 7, 2019 at 9:29

3 Answers 3

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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!

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

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

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