I'm fully comfortable with most series and even arithmetico–geometric sequence including n to any exponent if the geometric term is in the form of $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$, and so forth. However, I'm completely lost on how to calculate this sum by hand through term-by-term differentiation given the term is $\frac{5}{6}$. It would be appreciated if someone could help clear this up for me.
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https://math.stackexchange.com/questions/593996/how-to-prove-sum-n-0-infty-fracn22n-6/594019#594019 – lab bhattacharjee Oct 04 '18 at 04:24
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Thank you so much for this link. – Victor Petrescu Oct 04 '18 at 04:32
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Technically, not a geometric sequence. – Thomas Andrews Oct 04 '18 at 05:38
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Note generally that $$ \frac1{1-x} = f(x) = \sum_{k=0}^\infty x^k, \quad \forall x \in (-1,1). $$ Thus, $$ (1-x)^{-2} = f'(x) = \sum_{k=1}^\infty kx^{k-1}, $$ hence you also have $$ x(1-x)^{-2} = x\sum_{k=1}^\infty kx^{k-1} = \sum_{k=1}^\infty kx^k, $$ and now you can differentiate again to get $\sum k^2 x^k$ after another multiplication by $x$...
Since exchanging sum and differentiation is valid within the radius of convergence, all manipulations will hold for all $x \in (-1,1)$.
gt6989b
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Hint \begin{align*} \frac{1}{1-x}&=1+x+x^2+\dotsb=\sum_{k=0}^{\infty}x^k &&(\text{ for } |x| <1 )\\ \frac{d}{dx}(1-x)^{-1}=(1-x)^{-2}&=1+2x+3x^2\dotsb=\sum_{k=1}^{\infty}kx^{k-1}\\ x(1-x)^{-2}&=x+2x^2+3x^3\dotsb=\sum_{k=1}^{\infty}kx^{k} \end{align*} Now take derivative.
Anurag A
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Consider $$S=\sum_{n=1}^\infty n^2 x^{n-1}=\sum_{n=1}^\infty [n(n-1)+n] x^{n-1}=x\sum_{n=1}^\infty n(n-1) x^{n-2}+\sum_{n=1}^\infty n x^{n-1}$$ that is to say $$S=x\left(\sum_{n=1}^\infty x^n\right)''+\left(\sum_{n=1}^\infty x^n\right)'$$ When finished, make $x=\frac 56$.
Claude Leibovici
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