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Find the limit

$$\lim\limits_{n \rightarrow \infty}\frac{1}{(n+1) \log (1+\frac{1}{n})}$$

Kamil Jarosz
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El Chapo
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    Please show some effort on your end. What have you tried so far? What are you confused by? This is not a site for us to do your homework on. – TomGrubb Jul 22 '15 at 14:17
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    We're happy to give guidance, just not outright solve your problem. What have you tried so far? Is there some part of the theory you've learned that you're not sure on and maybe we can help elucidate? Or are you just stuck as to how to start? – ShakesBeer Jul 22 '15 at 14:19
  • Learn how to apply L'Hopital's rule https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule – karakfa Jul 22 '15 at 14:35
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    Hint: $\ln(1+1/n)\sim 1/n$ – Michael Galuza Jul 22 '15 at 14:38

5 Answers5

5

$$\lim\limits_{n \rightarrow \infty}\frac{1}{(n+1) \log (1+\frac{1}{n})}=\lim\limits_{n \rightarrow \infty}\frac{1}{\log\bigg[ (1+\frac{1}{n})^{n+1}\bigg]}$$

Now, use $$\lim_n (1+\frac{1}{n})^{n+1} =e$$

jameselmore
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N. S.
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3

Notice, $$\lim_{n\to \infty}\frac{1}{(n+1)\log\left(1+\frac{1}{n}\right)}$$ $$=\lim_{n\to \infty}\frac{1}{n\left(1+\frac{1}{n}\right)\log\left(1+\frac{1}{n}\right)}$$ $$=\lim_{n\to \infty}\frac{\left(\frac{1}{n}\right)}{\left(1+\frac{1}{n}\right)\log\left(1+\frac{1}{n}\right)}$$ Let $\frac{1}{n}=t \implies t\to 0\ as \ n\to \infty$ $$=\lim_{t\to 0}\frac{t}{\left(1+t\right)\log\left(1+t\right)}$$ Now, applying L-hospital's rule for $\frac{0}{0}$ form $$=\lim_{t\to 0}\frac{\frac{d(t)}{dt}}{\frac{d}{dt}\left(\left(1+t\right)\log\left(1+t\right)\right)}$$ $$=\lim_{t\to 0}\frac{1}{\left(1+t\right)\frac{1}{(1+t)}+(1)\log\left(1+t\right)}$$ $$=\lim_{t\to 0}\frac{1}{1+\log(1+t)}$$ $$=\frac{1}{1+\log(1+0)}$$ $$=\frac{1}{1+0}$$ $$=\frac{1}{1}=\color{blue}{1}$$

Harish Chandra Rajpoot
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$$\lim_{n\to\infty} \frac1{(n+1)\ln(1+\frac1n)} =\lim_{n\to\infty} \frac{(n+1)^{-1}}{\ln(1+\frac1n)}$$ Using l'Hopital's rule $\big(\frac00\big)$: $$\lim_{n\to\infty}\frac{(n+1)^{-2}}{\frac{1}{n^2}\big(\frac{n}{n+1}\big)} = \lim_{n\to\infty}\frac{n}{n+1} = 1$$

jameselmore
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2

Set $1/n=h\implies h\to0^+$

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

$$=\lim_{h\to0^+}\dfrac h{(h+1)\ln(1+h)}$$

$$=\dfrac1{\lim_{h\to0^+}\dfrac{\ln(1+h)}h}\cdot\dfrac1{\lim_{h\to0^+}(1+h)}=?$$

lab bhattacharjee
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1

HINT: $$\frac{1}{(n+1)\log\left(1+1/n\right)}=\frac{1}{\log\left(1+1/n)^n(1+1/n)\right)}=\frac{1}{\log(1+1/n)^n+\log(1+1/n)}$$

Dr. Sonnhard Graubner
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