Infinite Sum...

Calculus Level 4

∑ n = 1 ∞ n 3 2 n + 1 \large \sum_{n=1}^{\infty} \dfrac{n^3}{2^{n+1}}

Evaluate the infinite sum above to 3 decimal places.


The answer is 13.000.

Vilakshan Gupta by Vilakshan Gupta , 19, India

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2 solutions

Guilherme Niedu
Apr 1, 2018

By geometric progression formula:

∑ n = 1 ∞ x n = x 1 − x \large \displaystyle \sum_{n=1}^{\infty} x^n = \frac{x}{1-x}

Differentiating both sides with respect to x x and multiplying it by x x on both sides:

∑ n = 1 ∞ n x n = x ( 1 − x ) 2 \large \displaystyle \sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2}

Differentiating both sides with respect to x x and multiplying it by x x on both sides again:

∑ n = 1 ∞ n 2 x n = x + x 2 ( 1 − x ) 3 \large \displaystyle \sum_{n=1}^{\infty} n^2 x^n = \frac{x + x^2}{(1-x)^3}

Differentiating both sides with respect to x x and multiplying it by x x on both sides again:

∑ n = 1 ∞ n 3 x n = x + 4 x 2 + x 3 ( 1 − x ) 4 \large \displaystyle \sum_{n=1}^{\infty} n^3 x^n = \frac{x + 4x^2 + x^3}{(1-x)^4}

Now:

S = ∑ n = 1 ∞ n 3 ( 1 2 ) n + 1 \large \displaystyle S = \sum_{n=1}^{\infty} n^3 \left ( \frac12 \right ) ^{n+1}

S = 1 2 ⋅ ∑ n = 1 ∞ n 3 ( 1 2 ) n \large \displaystyle S = \frac12 \cdot \sum_{n=1}^{\infty} n^3 \left ( \frac12 \right ) ^n

S = 1 2 ⋅ ( 1 2 ) + 4 ( 1 2 ) 2 + ( 1 2 ) 3 ( 1 2 ) 4 \large \displaystyle S = \frac12 \cdot \frac{ \left ( \frac12 \right ) + 4 \left ( \frac12 \right ) ^2 + \left ( \frac12 \right ) ^3}{ \left ( \frac12 \right ) ^4 }

S = 13 \color{#3D99F6} \boxed{ \large \displaystyle S = 13}

3 Helpful 0 Interesting 1 Brilliant 0 Confused

Consider the following:

S 0 = ∑ n = 1 ∞ 1 2 n + 1 = 1 4 ∑ n = 0 ∞ 1 2 n = 1 4 ( 1 1 − 1 2 ) = 1 2 \begin{aligned} S_0 & = \sum_{\color{#3D99F6}n=1}^\infty \frac 1{2^{n+1}} = \frac 14 \sum_{\color{#D61F06}n=0}^\infty \frac 1{2^n} = \frac 14\left(\frac 1{1-\frac 12}\right) = \frac 12\end{aligned}

S 1 = ∑ n = 1 ∞ n 2 n + 1 = ∑ n = 0 ∞ n 2 n + 1 = ∑ n = 0 ∞ n + 1 2 n + 2 = 1 2 ∑ n = 0 ∞ n + 1 2 n + 1 = 1 2 S 1 + 1 2 ∑ n = 0 ∞ 1 2 n + 1 = 1 2 S 1 + 1 4 ∑ n = 0 ∞ 1 2 n = 1 2 S 1 + S 0 = 2 S 0 = 1 \begin{aligned} S_1 & = \sum_{\color{#3D99F6}n=1}^\infty \frac n{2^{n+1}} = \sum_{\color{#D61F06}n=0}^\infty \frac n{2^{n+1}} = \sum_{\color{#D61F06}n=0}^\infty \frac {n+1}{2^{n+2}} = \frac 12 \sum_{\color{#D61F06}n=0}^\infty \frac {n+1}{2^{n+1}} \\ & = \frac 12 S_1 + \frac 12 \sum_{\color{#D61F06}n=0}^\infty \frac 1{2^{n+1}} = \frac 12 S_1 + \frac 14 \sum_{\color{#D61F06}n=0}^\infty \frac 1{2^n} \\ & = \frac 12 S_1 + S_0 = 2S_0 = 1 \end{aligned}

S 2 = ∑ n = 1 ∞ n 2 2 n + 1 = ∑ n = 0 ∞ n 2 2 n + 1 = ∑ n = 0 ∞ ( n + 1 ) 2 2 n + 2 = 1 2 ∑ n = 0 ∞ n 2 + 2 n + 1 2 n + 1 = 1 2 S 2 + S 1 + S 0 = 2 S 1 + 2 S 0 = 6 S 0 = 3 \begin{aligned} S_2 & = \sum_{\color{#3D99F6}n=1}^\infty \frac {n^2}{2^{n+1}} = \sum_{\color{#D61F06}n=0}^\infty \frac {n^2}{2^{n+1}} = \sum_{\color{#D61F06}n=0}^\infty \frac {(n+1)^2}{2^{n+2}} = \frac 12 \sum_{\color{#D61F06}n=0}^\infty \frac {n^2+2n+1}{2^{n+1}} \\ & = \frac 12 S_2 + S_1 + S_0 = 2S_1 + 2S_0 = 6S_0 = 3 \end{aligned}

S 3 = ∑ n = 1 ∞ n 3 2 n + 1 = ∑ n = 0 ∞ n 3 2 n + 1 = ∑ n = 0 ∞ ( n + 1 ) 3 2 n + 2 = 1 2 ∑ n = 0 ∞ n 3 + 3 n 2 + 3 n + 1 2 n + 1 = 1 2 S 3 + 3 2 S 2 + 3 2 S 1 + S 0 = 3 S 2 + 3 S 1 + 2 S 0 = 26 S 0 = 13 \begin{aligned} S_3 & = \sum_{\color{#3D99F6}n=1}^\infty \frac {n^3}{2^{n+1}} = \sum_{\color{#D61F06}n=0}^\infty \frac {n^3}{2^{n+1}} = \sum_{\color{#D61F06}n=0}^\infty \frac {(n+1)^3}{2^{n+2}} = \frac 12 \sum_{\color{#D61F06}n=0}^\infty \frac {n^3+3n^2+3n+1}{2^{n+1}} \\ & = \frac 12 S_3 + \frac 32 S_2 + \frac 32 S_1 + S_0 = 3S_2 + 3S_1 + 2S_0 = 26S_0 = \boxed{13} \end{aligned}

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