Calculate the sun of the infinite series: ∑n=0∞n4n

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Calculate the sun of the infinite series:  ∑n=0∞n4n

hmotans · Accepted Answer

Your sum is equal to  14+242+343+444+…+n4n+…  Call this sum S. Now subtract from S the sum  14+142+143+…  If we do it in the obvious way, term by term, we obtain  142+243+344+…  Note that this last sum is (14)S  Putting things together, and using your computation for 1+14+142+… (not quite, we start at 1/4) we get  S−13=S4  Solve for S. We find that S=49

RizerMix · Accepted Answer

A simple probabilistic approach: Let X be a geometric random variable with probability of success p, so that P(X=n)=(1−p)n−1p, n=1,2,3,... Then the expectation of X is E(X)=∑n=1∞nP(X=n)=∑n=1∞n(1−p)n−1p=p1−p∑n=1∞n(1−p)n Thus ∑n=1∞n(1−p)n=1−ppE(X) On the other hand, from E(X)=p⋅1+(1−p)(1+E(X)) we get E(X)=1p Finally, ∑n=1∞n(1−p)n=1−ppE(X)=1−pp2 Letting p=3/4 gives ∑n=1∞n4n=1/49/16=49

nick1337 · Accepted Answer

First you can write ∑n=1∞n4n=∑n=1∞∑m=1∞14n Note that the above summation is over all (m,n) with m≤n. So if you switch the order of summation you obtain ∑m=1∞∑n=m∞14n The inner sum is a geometric series with initial term 14m and ratio 14, so it sums to 14m1−14=4314m. So the overall sum is 43∑m=1∞14m The sum here is a geometric series that sums to 13, so your final answer is 43×13=49

Calculate the sun of the infinite series: \sum_{n=0}^\infty\frac{n}{4^n}

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