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Compute $T$ for $n>0$; $$T=\lim_{n \to \infty} \int_{\frac{1}{\sqrt [n+1] {n!}}}^{\frac{1}{\sqrt [n+1]{(n-1)!}}} \frac{n+1}{x(n+1)ln(x)+xln((n+1)!)} \,dx$$

My attempts;

Whatever I have tried was along the lines of this;

Apart from that,

$$T=\lim_{n \to \infty} \int_{\frac{1}{\sqrt [n+1] {n!}}}^{\frac{1}{\sqrt [n+1]{(n-1)!}}} \frac{1}{x\left[ln(x)+\frac{1}{n+1}ln((n+1)!)\right]} \,dx$$

Putting $\frac{1}{n+1}ln((n+1)!)=t$

$$T=\lim_{n \to \infty} \int_{\frac{1}{\sqrt [n+1] {n!}}}^{\frac{1}{\sqrt [n+1]{(n-1)!}}} \frac{1}{x\left[ln(x)+t\right]} \,dx$$

Ignoring the limit and the bounds as noise, you get;

$$T= \int \frac{1}{x\left[ln(x)+t\right]} \,dx$$

Substitute $ln(x)=u$ and simplifying,

$$T=ln(ln(x)+t)$$

$$T=ln\left[ln(x)+\frac{1}{n+1}ln((n+1)!)\right]_{\frac{1}{\sqrt [n+1] {n!}}}^{\frac{1}{\sqrt [n+1]{(n-1)!}}}$$

$$T=ln\left[ln\left(\frac{1}{\sqrt [n+1]{(n-1)!}}\right)+\frac{1}{n+1}ln((n+1)!)\right]-ln\left[ln\left(\frac{1}{\sqrt [n+1] {n!}}\right)+\frac{1}{n+1}ln((n+1)!)\right]$$

$$T=\lim_{n \to \infty} ln\left[ln\left(\frac{1}{\sqrt [n+1]{(n-1)!}}\right)+\frac{1}{n+1}ln((n+1)!)\right]-ln\left[ln\left(\frac{1}{\sqrt [n+1] {n!}}\right)+\frac{1}{n+1}ln((n+1)!)\right]$$

But I hit a dead-end here. Although it is unnecessary (I think), I reading about convergence tests just to check its convergence but could not apply it to conclude anything. If my work was right, how do I proceed from here? I am thinking use of Stirling approximation.

Also what is the best method to approach this/such kind of limits?

Amrut Ayan
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1 Answers1

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What you did seems correct. To continue

$\ln \left[\ln\left( \frac{1}{\sqrt[n+1]{(n-1)!}}\right) + \frac{1}{n+1} \ln((n+1)!)\right] - \ln \left[\ln\left( \frac{1}{\sqrt[n+1]{n!}}\right) + \frac{1}{n+1} \ln((n+1)!)\right]$ $= \ln\left(\frac{\ln\left( \frac{1}{\sqrt[n+1]{(n-1)!}}\right) + \frac{1}{n+1} \ln((n+1)!)}{\ln\left( \frac{1}{\sqrt[n+1]{n!}}\right) + \frac{1}{n+1} \ln((n+1)!)}\right)=\ln\left(\frac{\ln\left( \frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n+1]{(n-1)!}}\right) }{\ln\left( \frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n+1]{n!}}\right) }\right) = \ln\left(\frac{\ln\left( {\sqrt[n+1]{\frac{(n+1)!}{(n-1)!}}}\right) }{\ln\left( {\sqrt[n+1]{\frac{(n+1)!}{n!}}}\right) }\right)$ $=\ln\left(\frac{\frac{1}{n+1}\ln\left( n(n+1)\right) }{\frac{1}{n+1}\ln\left( n+1\right) }\right)=\ln\left(\frac{\ln\left( n(n+1)\right) }{\ln\left( n+1\right) }\right)=\ln\left(1+\frac{\ln\left( n\right) }{\ln\left( n+1\right) }\right)$

$\Rightarrow T = \lim_{n\rightarrow \infty}{\ln\left(1+\frac{\ln\left( n\right) }{\ln\left( n+1\right) }\right)}=\ln(2)$

Mehdi
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    @ayan This post is most of the proof...The limit was not taken until the very end of the manipulations. How formally you wish to take it was not specified by you at all. It comes across as you couldn't be bothered to apply basic properties of limits yourself after all this work was already done for you. – Ninad Munshi Jul 28 '23 at 20:25
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    @ayan Doesn't the last line show you what value the limit equals? – Accelerator Jul 29 '23 at 00:52