Can someone explain me how to find sum of next series:
$\sum_{n=1}^\infty n^4 \tan^{n-1}(x)$
Thanks for answers in advance.
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Well Maple gives me the answer, I'd also be interested in knowing how that word. – Matt Samuel Nov 13 '14 at 19:35
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@Matt What is the answer ? – Peter Nov 13 '14 at 19:46
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$$\frac{-\tan^3(x)-11\tan^2(x)-11\tan(x)-1}{(\tan(x)-1)^5}$$ Obviously this only works if $|\tan(x)|<1$. – Matt Samuel Nov 13 '14 at 19:51
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1Hint : Consider the sum $$\sum_{n=1}^{\infty} n^4t^{n-1}$$ – Peter Nov 13 '14 at 19:54
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Another hint : Consider the derivation of $n^4t^{n-1}$ with respect to $t$ – Peter Nov 13 '14 at 19:57
2 Answers
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We can write $n^4$ as a sum of combinatorial polynomials of degree $4$ or less: $$ n^4=24\binom{n}{4}+36\binom{n}{3}+14\binom{n}{2}+\binom{n}{1}\tag{1} $$ Next, for $|x|\lt1$, we can use the Generalized Binomial Theorem and negative binomial coefficients to show $$ \begin{align} \sum_{n=k}^\infty\binom{n}{k}x^{n-1} &=\sum_{n=k}^\infty\binom{n}{n-k}x^{n-1}\\ &=\sum_{n=k}^\infty\binom{-k-1}{n-k}(-1)^{n-k}x^{n-1}\\ &=\sum_{n=0}^\infty\binom{-k-1}{n}(-1)^{n}x^{n+k-1}\\ &=\frac{x^{k-1}}{(1-x)^{k+1}}\tag{2} \end{align} $$ Therefore, using $(1)$ and $(2)$, we get $$ \begin{align} \sum_{n=1}^\infty n^4x^{n-1} &=\frac{24x^3}{(1-x)^5}+\frac{36x^2}{(1-x)^4}+\frac{14x}{(1-x)^3}+\frac1{(1-x)^2}\\ &=\frac{1+11x+11x^2+x^3}{(1-x)^5}\tag{3} \end{align} $$ Now, just substitute $x\mapsto\tan(x)$.
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Hint: $~\displaystyle\sum_{n=0}^\infty t^n~=~?~$ Now, differentiate both sides with respect to t, and notice what happens... :-)
Lucian
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