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$$\frac12\ln2+\frac1{2^2}\ln2^2+\frac1{2^3}\ln2^3+\cdots+\frac1{2^m}\ln2^m+\cdots$$ The answer says this series should sum to 4. Could someone please help how to get the correct result? I am confused which formulas to use and how.

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2  
Is that $\log(2^m)$ or $(\log 2)^m$? –  WimC Nov 6 '12 at 16:50
    
Use geometric series! –  Chasky Nov 6 '12 at 16:52
    
It's log(2^m), only the two goes to the power. –  red8 Nov 6 '12 at 17:06

3 Answers 3

up vote 1 down vote accepted

$m$th term is $\frac 1{2^m}\ln2^m=(\ln 2){\frac m{2^m}}$

$$\sum_{1\le r<\infty} \ln 2^{\frac r{2^r}}=(\ln2){\sum_{1\le r<\infty}\frac m{2^m}}$$

Let $S={\sum_{1\le r<\infty}\frac m{2^m}}=\frac 12+\frac 2{2^2}+\frac 3{2^3}+\cdots$ (This is a Arithmetico-geometric series)

So, dividing either sides by $2,\frac S 2=\frac 1{2^2}+\frac 2{2^3}+\frac 3{2^4}+\cdots$

On subtraction, $\frac S 2=\frac 12+\frac 1{2^2}+\frac 1{2^3}+\cdots$

$\frac S 2=\frac {\frac 12}{1-\frac 12}$ (Using infinite geometric Progression with common ratio$=\frac 12<1$)

So, $S=2$

$$\sum_{1\le r<\infty} \ln 2^{\frac r{2^r}}=(\ln2){\sum_{1\le r<\infty}\frac m{2^m}}=2\ln2=\ln 4$$

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First note that $\ln 2^m=m\ln 2$. Then we see that $$\sum_{k=1}^\infty \frac1{2^k}\ln 2^k = \ln2\cdot\sum_{k=1}^\infty \frac k{2^k}$$ Next observe that $$\sum_{k=1}^\infty \frac k{2^k}= \sum_{k=1}^\infty \frac12\frac {k-1}{2^{k-1}}+\sum_{k=1}^\infty \frac 1{2^k}=\frac12\sum_{k=1}^\infty \frac k{2^k}+1.$$ Hence (once you agree that $\sum_{k=1}^\infty \frac k{2^k}$ converges) $$\sum_{k=1}^\infty \frac1{2^k}\ln 2^k=2\ln2.$$

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Hint: $\sum\limits_{n\geqslant1}nx^n=x/(1-x)^2$ for every $|x|\lt1$. (And the answer is not $4$.)

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