How to determine the sum of series $\sum_{k=1}^{\infty} \frac{k^4}{3^k}$
The series converges and I found that:
$\sum_{k=1}^{\infty} \frac{k^4}{3^k} = \frac{1}{3} + \frac{16}{9} + 3 + \frac{256}{81} ...$
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olava
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4Hint: apply $x\frac{d}{dx}$ four times to $\sum_{k\ge1}x^k=\frac{x}{1-x}$, then set $x=\frac13$. – J.G. Nov 11 '22 at 22:10
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3Does this answer your question? Why do expressions of the form $\sum\limits^\infty_{n=1} \frac{n^k}{3^n}$ sum 'nicely'?. See also Summation of series $\sum_{n=1}^\infty \frac{n^a}{b^n}$? and How to compute sums like $\sum_{n=1}^{\infty} \frac{n^a}{b^n} $. – Anne Bauval Nov 11 '22 at 22:27
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No I shouldn't use derivatives to solve that – olava Nov 11 '22 at 22:48
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Did someone not tell you to use derivatives? That’s the only method I know. – Lubin Nov 11 '22 at 22:54
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Could you explain how to use derivatives in this case? – olava Nov 11 '22 at 22:56
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(i) Use a recursive method or Stirling numbers(basically the same thing) to write $k^4$ as a lonear combination of falling powers $$ak^{(4)}+bk^{(3)}+ck^{(2)}+dk^{(1)}$$(ii) write $$(1/3)^k=(-3/2)\Delta(1/3)^k$$(iii)Sum the terms from $k=1 \space \text {to} k=m$ by applying summation by parts 4 times to reduce to a geometric series. $$\text{(iv)Take the limit as } m \rightarrow \infty$$
P. Lawrence
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