0
$\begingroup$

I have searched here & google but find no clues about how to find it out. This is clearly alternating series, if it is convergent then we have in our hand Leibnitz test, Dirichlet test, Abel test & something like this. In fact we are able to show the given series is absolutely convergent then we are done. But if divergent, then how to show a alternating series is divergent. If we are able to show $f(n)=\log(n+1)/(n+1)$ is decreasing then also we are done. But is the $f(n)$ actually decreases? If so then how? Please help me to get out of this. Thanks in advance.

$\endgroup$
  • $\begingroup$ take a derivative of f(x) = log(x+1)/(x+1) and show it is negative for all x in some interval (c, infinity). $\endgroup$ – Leonard Blackburn Apr 26 at 16:00
1
$\begingroup$

You want to show that, from some point on, $$ \frac{\log(n+1)}{n+1}<\frac{\log n}{n} $$ which is equivalent to $$ (n+1)^n<n^{n+1} $$ or $$ \frac{(n+1)^n}{n^n}<n $$ Since this is the same as $$ \left(1+\frac{1}{n}\right)^n<n $$ you know that this eventually holds (actually for $n\ge3$, but it's irrelevant), because the limit of the left-hand side is $e$.

If you want to prove that without limits, you can consider https://math.stackexchange.com/a/1868235/62967. Basically, \begin{align} \left(1+\frac{1}{n}\right)^n &= 1+1+\frac{1}{2!}\left(1-\frac{1}{n}\right)+ \frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\\ &\qquad+\dots+ \frac{1}{n!}\left(1-\frac{1}{n}\right)\dotsm\left(1-\frac{n-1}{n}\right) \\[6px] &< 1+1+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{n!} \\[6px] &< 1+1+\frac{1}{2}+\frac{1}{2^2}+\dots+\frac{1}{2^n} \\[6px] &<1+2=3 \end{align}

On the other hand, showing that $f(x)=\frac{\log x}{x}$ is eventually decreasing is much easier: $$ f'(x)=\frac{1-\log x}{x^2} $$ which is negative as soon as $x>e$.

$\endgroup$
  • $\begingroup$ 's sir actually I want this, but sir I was unable to make comment about (1+1/n)^n<n, sir please tell me how we can say this? $\endgroup$ – user639336 Apr 26 at 16:40
  • $\begingroup$ @user639336 The limit of $(1+1/n)^n$ is $e$, which is less than $3$. $\endgroup$ – egreg Apr 26 at 16:46
  • $\begingroup$ sir I want to show simply that f(n+1)<=f(n), for large n, from your statement how it follows? You have just shown for large n (1+1/n)^n<3,please tell me that. $\endgroup$ – user639336 Apr 26 at 17:13
  • $\begingroup$ @user639336 Since $(1+1/n)^n<3$, for every $n>0$, we have that $(1+1/n)^n<n$ when $n\ge3$. $\endgroup$ – egreg Apr 26 at 17:14
  • $\begingroup$ sir thank you & sorry that was infront of me but couldn't see. Thank you sir. By considering function I'm done. $\endgroup$ – user639336 Apr 26 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.