This is the problem:
$$\sum_{n=2}^\infty \frac{n(n-1)}{2^n}$$
How can we write the exact sum of the series and determine if it is converges or not?
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Solution 1: Take two derivatives of $\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ and evaluate at $x=\frac{1}{2}$. Solution 2: Multiply its partial sums by $\left(1-\frac{1}{2}\right)^2$. – logarithm May 18 '19 at 19:58
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This question is pretty similat. – logarithm May 18 '19 at 20:00
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3 Answers
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Convergence is trivial from either the ratio or root tests.
To evaluate the sum, write $f(x):=\sum (x/2)^n=1/(1-x/2)$. Now notice that your sum is $f’’(1)$. In other words take two derivatives of the right side and plug in $x=1$. The answer should be 4.
Alex R.
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Hint:
$$\sum_{n=2}^\infty n(n-1)x^n=x^2\Bigl(\sum_{n=2}^\infty n(n-1)x^{n-2}\Bigr)=x^2\Bigl(\sum_{n=0}^\infty x^{n}\Bigr)^{\!''}.$$
Bernard
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@cansomeonehelpmeout: Yes, of course. Sorry for the mistyping. It's fixed now. – Bernard May 18 '19 at 20:06
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$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$ $$(\frac{1}{1-x})''=\sum_{n=2}^\infty n(n-1)x^{n-2}$$ $$x^2(\frac{1}{1-x})''=\sum_{n=2}^\infty n(n-1)x^{n}$$ $$\frac{2x^2}{(1-x)^3}=\sum_{n=2}^\infty n(n-1)x^{n}$$ now let $x=\frac{1}{2}$
E.H.E
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