Question

The numerator of the first term above should look like this:  

\(\ln[(1+1/n)^n (n+1)]\)

i.e. all the terms containing n are part of the argument of the logarithm function.

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0 mins ago

I entered it in W/A engine and it gave this result. Check and see if it is my mistake in entering it,

or check to see if you made a mistake:

sum_(n=2)^∞ (log((1 + 1/n)^n (n + 1)))/(log(n^n) log^(n + 1)(n + 1))≈1.27891

Well i didn't make this exercise, i just have to solve it so i don't know if it's wrong or not. What i do know is i have written it correctly.

I talked about this yesterday but my post disappeared. :(

I think that guests Wolfram Alpha input is incorrect.  

I have just copied and pasted guests input into Wolfram|Alpha and it was not entered correctly and Wolfram Alpha said it was equal to infinity...

https://www.wolframalpha.com/input/?i=sum_(n%3D2)%5E%E2%88%9E+(log((1+%2B+1%2Fn)%5En+(n+%2B+1)))%2F(log(n%5En)+log%5E(n+%2B+1)(n+%2B+1))

I put it into Wofram Alpha too.  

https://www.wolframalpha.com/input/?i=sum_(n%3D2)%5E%E2%88%9E+((log((1+%2B+1%2Fn)%5En+(n+%2B+1)))%2F(log(n%5En)+(log((n+%2B+1)%5E(n+%2B+1))))

I got no final answer but I was told that it was convergent and the graph indicated the answer was about 0.7  

(1/(2*log(2,e))) = 0.7213475204444817

So   It seemed that    1/(2ln2) is  likely to be the answer.

I personally have no idea how to do it though.  

I do not know how to do it :((

If you were a member then you would be able to private message Heureka or Alan (with the address of your question included) and ask them politely if they would be able to help you .:)

Like so:

\(f(n)=\frac{\ln(1+1/n)^n(n+1)}{\ln(n^n)\ln(n+1)^{n+1}}\\\\ f(n)=\frac{(n+1)\ln(n+1)-n\ln(n)}{n\ln(n)(n+1)\ln(n+1)}\\\\ f(n)=\frac{1}{n\ln(n)}-\frac{1}{(n+1)\ln(n+1)}\\\\ \Sigma_{n=2}^\infty f(n)=\frac{1}{2\ln2}-\frac{1}{3ln3}+\frac{1}{3ln3}-\frac{1}{4ln4}+\frac{1}{5ln5}-\frac{1}{5ln5}+...\\\\ \Sigma_{n=2}^\infty f(n)=\frac{1}{2ln2}\)

i.e. the only term that doesn't get cancelled out is the first one.

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Hi ! I have to show that
\(\sum \limits_{n=2}^{\infty} \left( \dfrac { \ln \left[(1+ \frac {1} {n})^n\cdot (n+1) \right]} { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right]} \right)= \dfrac {1} {2\cdot \ln(2)}\)
Can anyone help ? Thank you in advance.

\(\begin{array}{|rcll|} \hline && \sum \limits_{n=2}^{\infty} \dfrac { \ln \left[(1+ \frac {1} {n})^n\cdot (n+1) \right]} { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right]} \\\\ &=&\sum \limits_{n=2}^{\infty} \dfrac { \ln \left[(\frac {n+1} {n})^n\cdot (n+1) \right]} { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right]} \\\\ &=&\sum \limits_{n=2}^{\infty} \dfrac { \ln \left[ \frac {(n+1)^n} {n^n}\cdot (n+1) \right]} { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right]} \\\\ &=&\sum \limits_{n=2}^{\infty} \dfrac { \ln [ \frac {(n+1)^{n+1}} {n^n} ]} { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right]} \\\\ &=&\sum \limits_{n=2}^{\infty} \dfrac { \ln \left[(n+1)^{n+1} \right]- \ln(n^n) } { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right] } \\\\ &=&\sum \limits_{n=2}^{\infty} \left( \dfrac { \ln \left[(n+1)^{n+1} \right] } { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right] } - \dfrac { \ln(n^n) } { \ln(n^n)\cdot \ln \left[(n+1)^{n+1} \right] } \right) \\\\ &=&\sum \limits_{n=2}^{\infty} \left( \dfrac { 1 } { \ln(n^n) } - \dfrac { 1 } { \ln \left[(n+1)^{n+1} \right] } \right) \\\\ &=&\sum \limits_{n=2}^{\infty} \dfrac { 1 } { \ln(n^n) } -\sum \limits_{n=2}^{\infty} \dfrac { 1 } { \ln \left[(n+1)^{n+1} \right] } \\\\ &=&\sum \limits_{n=1}^{\infty} \dfrac { 1 } { \ln[(n+1)^{n+1}] } -\sum \limits_{n=2}^{\infty} \dfrac { 1 } { \ln \left[(n+1)^{n+1} \right] } \\\\ &=& \dfrac{1}{\ln(2^2)} + \sum \limits_{n=2}^{\infty} \dfrac { 1 } { \ln[(n+1)^{n+1}] } -\sum \limits_{n=2}^{\infty} \dfrac { 1 } { \ln \left[(n+1)^{n+1} \right] } \\\\ &=& \dfrac{1}{\ln(2^2)} \\\\ &=& \dfrac{1}{2\cdot \ln(2)} \\ \hline \end{array}\)

Thank you to everyone who got involved ! I'll consider creating an account, much appreciated the help from everyone ! :)