How to prove: ∑n=0∞n22n=6

Question

How to prove:  ∑n=0∞n22n=6

Steve Hirano · Accepted Answer

First observe that the sum converges (by, say, the root test).
We already know that

R = \sum_{n = 0}^{\infty} \frac{1}{2^{n}} = 2

Let S be the given sum. Then

S = 2 S - S = \sum_{n = 0}^{\infty} \frac{2 n + 1}{2^{n}}

Now use the same trick to compute

T = \sum_{n = 0}^{\infty} \frac{n}{2^{n}}

we have

T = 2 T - T = \sum_{n = 0}^{\infty} \frac{1}{2^{n}} = R = 2

.
Hence

S = 2 T + R = 6

One can continue like this to compute

X = \sum_{n = 0}^{\infty} \frac{n^{3}}{2^{n}}

. We have

X = 2 X - X = \sum_{n = 0}^{\infty} \frac{3 n^{2} + 3 n + 1}{2^{n}} = 3 S + 3 T + R = 26

. Sums with larger powers can be computed in the same way.

Annie Gonzalez · Accepted Answer

If Calculus is allowed, using infinite geometric series formula for |r|&amp;lt;1  ∑0≤n&amp;lt;∞rn=11−r  Differentiating wrt r  ∑0≤n&amp;lt;∞nrn−1=1(1−r)2  ⇒∑0≤n&amp;lt;∞nrn=r(1−r)2=1−(1−r)(1−r)2=1(1−r)2−1(1−r)  Differentiating wrt r  ⇒∑0≤n&amp;lt;∞n2rn−1=2(1−r)3−1(1−r)2  ⇒∑0≤n&amp;lt;∞n2rn=2r(1−r)3−r(1−r)2  Here r=12

alenahelenash · Accepted Answer

Consider 11−z=∑n=0∞zn Differentiating and multiplying by z, one has z(1−z)2=z∑n=1∞nzn−1=∑n=1∞nzn Repeating the above process, ∑n=1∞n2zn=z(1+z)(1−z)3 Now plug iz z=1/2

How to prove: \sum_{n=0}^\infty\frac{n^2}{2^n}=6

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