So, yes, I could not do anything except observing that in the denominators, there is a geometric progression and in the numerator, $1^4+2^4+3^4+\cdots$.
Edit: I don't want the proof of it for divergence or convergence only the sum.
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Aditya Agarwal
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If you only want the result, go to WA... – user37238 Sep 01 '15 at 15:12
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With the method of obtaining the result. – Aditya Agarwal Sep 01 '15 at 15:15
4 Answers
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$$\sum_{n\geq 1}\frac{e^{nx}}{4^n} = \frac{e^x}{4-e^x} $$ since it is the sum of a geometric progression. Differentiating four times with respect to $x$, $$\sum_{n\geq 1}\frac{n^4}{4^n}e^{nx} = \frac{256 e^x+704 e^{2 x}+176 e^{3 x}+4 e^{4 x}}{(4-e^x)^5}$$ and by evaluating at $x=0$,
$$\sum_{n\geq 1}\frac{n^4}{4^n}= \color{red}{\frac{380}{81}}.$$
Jack D'Aurizio
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Let
$$f(x) = \sum_{n=1}^{\infty} x^n = \frac{x}{1-x} $$
Then the sum is
$$\left [\left (x \frac{d}{dx} \right )^4 f(x)\right ]_{x=1/4}$$
Ron Gordon
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1And didn't quite understand the notation in the second expression? – Aditya Agarwal Sep 01 '15 at 15:15
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I think OP deserves more explanation as for why this expression gives the sum. – Wojowu Sep 01 '15 at 15:15
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@AdityaAgarwal: sorry for the typo, yes. And $$\left ( x \frac{d}{dx} \right )^2 f(x) = x \frac{d}{dx} \left ( x \frac{df}{dx} \right ) $$ carry on to $n=4$ – Ron Gordon Sep 01 '15 at 15:16
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1@AdityaAgarwal You differentiate $f$ then multiply it by $x$. So applying $x\frac{\mathrm d}{\mathrm d x}$ to $\sum x^n$ gives us $\sum nx^n$ Repeating it $4$ times gives $\sum n^4 x^n$ – Kitegi Sep 01 '15 at 15:17
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1@AdityaAgarwal That's why we need to multiply by $x$. That way, we get $x*nx^{n-1} = nx^n$ – Kitegi Sep 01 '15 at 15:24
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But if you try it on WA, then answer comes out to be $\frac{-380}{81}$ (according to your method). But real answer is the positive form of this number. – Aditya Agarwal Sep 01 '15 at 15:28
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@AdityaAgarwal No, you've probably made an error in entering it into Wolfram Alpha, or missed a $-$ sign in reading the output. – Erick Wong Sep 01 '15 at 15:39
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I've always done these kinds of sums by hand and I've never heard of this technique! – Eemil Wallin Sep 01 '15 at 15:41
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Consider that if we take a simple power series: $$f(x)=\sum_{n=0}^\infty a_n x^n$$ we can differentiate term-by-term to find a power series for its derivative, giving $$\frac{d}{dx}\left[f\right]=\sum_{n=0}^\infty na_n x^{n-1}$$To restore the power of $x$ to $n$ from $n-1$, consider we can multiply throughout by $x$ to yield $$x\cdot\frac{d}{dx}\left[f\right]=\sum_{n=0}^\infty na_n x^n$$ In general, if we have a power series $\sum a_n x^n$ we can differentiate and multiply by $x$ to give the power series $\sum na_n x^n$, so to get $\sum n^4 a_n x^n$ we can simply repeat this process another three times: $$\begin{align*}&f(x)=\sum_{n=0}^\infty a_n x^n\\\rightarrow\ &x \frac{df}{dx}=\sum_{n=0}^\infty na_n x^n\\\rightarrow\ &x\frac{d}{dx}\left(x\frac{df}{dx}\right)=\sum_{n=0}^\infty n^2 a_n x^n\\\dots\\\rightarrow\ &x\frac{d}{dx}\left(x\frac{d}{dx}\left(x\frac{d}{dx}\left(x\frac{df}{dx}\right)\right)\right)=\sum_{n=0}^\infty n^4 a_n x^n\end{align*}$$
To condense the expression on the left, we can treat this operation -- differentiate and then multiply by $x$ -- as its own operator $x\cdot\frac{d}{dx}$. Using power notation to denote repeated application $(x\cdot\frac{d}{dx})^{n}f=x\frac{d}{dx}(\dots (x\frac{df}{dx}))$ we have $$\left(x\cdot\frac{d}{dx}\right)^4f=\sum_{n=0}^\infty n^4 a_n x^n$$
Now consider $a_n=1$, giving the geometric series: $$f(x)=\sum_{n=0}^\infty x^n$$We know a closed-form for $f$, namely $$f(x)=\frac1{1-x}$$ We can then differentiate this expression and multiply by $x$, etc. the right number of times to get a closed-form expression for $\sum n^4 x^n$.
Once we have a closed-form expression for $\left(x\cdot\frac{d}{dx}\right)^4 f$, provided that the series converges for $x=\frac14$, we can evaluate said function at $x=\frac14$ to evaluate the series in your post: $$\left(x\cdot\frac{d}{dx}\right)^4 f=\sum_{n=0}^\infty n^4 x^n\\\left[\left(x\cdot\frac{d}{dx}\right)^4 f\right]_{x=\frac14}=\sum_{n=0}^\infty \frac{n^4}{4^n}$$
obataku
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\begin{align} n^4 = {} & \hphantom{{}+{}} An(n-1)(n-2)(n-3) \\ & {} + B n(n-1)(n-2) \\ & {} + C n(n-1) \\ & {} + D n \end{align} Find $A,B,C,D$. Then \begin{align} n^4 x^n & = An(n-1)(n-2)(n-3) x^n + \cdots \\[10pt] & = A\frac{d^4}{dx^4} x^n + B \frac{d^3}{dx^3} x^n + \cdots \end{align}
Then add over all values of $n$, and then find the derivatives.
Finally, apply this to the case where $x=\dfrac 1 4$.
