Infinite Sum Practice (3)

Calculus Level 5

Evaluate

S = 2 3 ⋅ e 3 + 4 5 ⋅ e 5 + 6 7 ⋅ e 7 + 8 9 ⋅ e 9 + ⋯ S=\dfrac{2}{3 \cdot e^3}+\dfrac{4}{5 \cdot e^5}+\dfrac{6}{7 \cdot e^7}+\dfrac{8}{9 \cdot e^9}+\cdots


The answer is 0.0395.

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Pratik Shastri by Pratik Shastri , 35, India

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1 solution

Ronak Agarwal
Nov 2, 2014

We have S = ∑ n = 1 ∞ 2 n 2 n + 1 e − ( 2 n + 1 ) \large \displaystyle S=\sum _{ n=1 }^{ \infty }{ \frac { 2n }{ 2n+1 } { e }^{ -(2n+1) } }

Rewriting it we have :

S = ∑ n = 1 ∞ e − ( 2 n + 1 ) − ∑ n = 1 ∞ e − ( 2 n + 1 ) 2 n + 1 = S 1 − S 2 \large \displaystyle S=\sum _{ n=1 }^{ \infty }{ { e }^{ -(2n+1) } } -\sum _{ n=1 }^{ \infty }{ \frac { { e }^{ -(2n+1) } }{ 2n+1 } }={S}_{1}-{S}_{2}

S 1 {S}_{1} is a infinite G.P. which can be found easily and it comes out to be :

S 1 = 1 e ( e 2 − 1 ) \large {S}_{1}=\frac{1}{e({e}^2-1)}

We know that :

∑ n = 1 ∞ x 2 n = x 2 1 − x 2 \large \displaystyle \sum _{ n=1 }^{ \infty }{ { x }^{ 2n } } =\frac { { x }^{ 2 } }{ 1-{ x }^{ 2 } }

Integrating both sides we get :

∑ n = 1 ∞ x 2 n + 1 2 n + 1 = 1 2 l n ( 1 + x 1 − x ) − x + C \large \displaystyle \sum _{ n=1 }^{ \infty }{ \frac { { x }^{ 2n+1 } }{ 2n+1 } } =\frac { 1 }{ 2 } ln(\frac { 1+x }{ 1-x } )-x+C

Put x = 0 x=0 to get C = 0 C=0 hence :

∑ n = 1 ∞ x 2 n + 1 2 n + 1 = 1 2 l n ( 1 + x 1 − x ) − x \large \displaystyle \sum _{ n=1 }^{ \infty }{ \frac { { x }^{ 2n+1 } }{ 2n+1 } } =\frac { 1 }{ 2 } ln(\frac { 1+x }{ 1-x } )-x

Put x = 1 e x=\frac{1}{e} to get S 2 {S}_{2} as :

S 2 = ∑ n = 1 ∞ x 2 n + 1 2 n + 1 = 1 2 l n ( e + 1 e − 1 ) − 1 e \large \displaystyle {S}_{2}=\sum _{ n=1 }^{ \infty }{ \frac { { x }^{ 2n+1 } }{ 2n+1 } } =\frac { 1 }{ 2 } ln(\frac { e+1 }{ e-1 } )-\frac { 1 }{ e }

Finally we get :

S = 1 e ( e 2 − 1 ) − ( 1 2 l n ( e + 1 e − 1 ) − 1 e ) \large S=\frac { 1 }{ e({ e }^{ 2 }-1) } -(\frac { 1 }{ 2 } ln(\frac { e+1 }{ e-1 } )-\frac { 1 }{ e } )

Simplifying this we get :

S = e ( e 2 − 1 ) − 1 2 l n ( e + 1 e − 1 ) \large S=\frac { e }{ ({ e }^{ 2 }-1) } -\frac { 1 }{ 2 } ln(\frac { e+1 }{ e-1 } )

15 Helpful 0 Interesting 0 Brilliant 0 Confused

did the exact same save that I came to the fact that ∑ n = 1 ∞ x 2 n + 1 2 n + 1 = 1 2 ln ⁡ ( 1 + x 1 − x ) − x \sum _{ n=1 }^{ \infty }{ \frac { { x }^{ 2n+1 } }{ 2n+1 } } \quad =\quad \frac { 1 }{ 2 } \ln { (\frac { 1+x }{ 1-x } } )\quad -\quad x in a different manner. I did it like this -

∑ n = 1 ∞ x n n = − ln ⁡ ( 1 − x ) \sum _{ n=1 }^{ \infty }{ \frac { { x }^{ n } }{ n } } \quad =\quad -\ln { (1-x } )\quad after integrating this - ∑ n = 0 ∞ x n = 1 1 − x \sum _{ n=0 }^{ \infty }{ { x }^{ n } } \quad =\quad \frac { 1 }{ 1-x } \quad

Also, ∑ n = 0 ∞ x 2 n + 1 − ∑ n = 1 ∞ x 2 n = ln ⁡ ( 1 + x ) \sum _{ n=0 }^{ \infty }{ { x }^{ 2n+1 } } -\sum _{ n=1 }^{ \infty }{ { x }^{ 2n } } \quad =\quad \ln { (1+x) } \quad

Adding both these and then some bashing.

Kartik Sharma - 6 years, 6 months ago

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Of course yours is better! And yes nice question!

Kartik Sharma - 6 years, 6 months ago

This is how I originally thought, while making the question that is..

Pratik Shastri - 6 years, 6 months ago

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Hmm but Ronak's way is better, I think. What say?

Kartik Sharma - 6 years, 6 months ago

You might like to use \displaystyle before \sum.

  • ∑ 1 2 n = \sum_{1}^{2} n= \sum

  • ∑ 1 2 n = \displaystyle\sum_{1}^{2} n= \displaystyle\sum

Pranjal Jain - 6 years, 4 months ago

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