How should I rewrite the sequence so that we have a form that we can easily calculate the limit?
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1$\sum_{a=1}^{\infty}\sum_{b=a}^{\infty}\frac1{2^b}$ – Empy2 Feb 23 '16 at 04:06
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Hint:
Start with: $$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$ for $|x|<1$.
Now, differentiate both sides, we get that: $$\sum_{n=0}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$$ Therefore, $$\sum_{n=0}^\infty nx^{n}=\frac{x}{(1-x)^2}$$ Plug in $x=\frac{1}{2}$
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Thanks so much! This is really clever, I didn't know you can differentiate a series like a normal function. Also, is it necessary to find the limit to conclude that it is convergent? Or we can use a simpler method to quickly determine if it is convergent? – lsy Feb 23 '16 at 04:18
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