Simple Problem on Limits!

Calculus Level 4

L = lim ⁡ n → ∞ ( 1 3 5 + 2 4 5 + 3 5 5 + … + n ( n + 2 ) 5 ) \large{L=\lim_{n \to \infty} \left( \dfrac1{3^5} + \dfrac2{4^5} + \dfrac3{5^5} + \ldots + \dfrac{n}{(n+2)^5} \right)}

Choose the correct option.

You are given that ζ ( 5 ) ≈ 1.03692 \zeta(5) \approx 1.03692 .

0.1 < L ≤ 1 0.1 < L \leq 1 L ≤ 0.01 L \leq 0.01 L > 1 L > 1 0.01 < L ≤ 0.1 0.01 < L \leq 0.1

Satyajit Mohanty by Satyajit Mohanty , 24, India

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2 solutions

Maggie Miller
Jul 23, 2015

∑ k = 1 ∞ k ( k + 2 ) 5 = ∑ k = 1 ∞ k + 2 ( k + 2 ) 5 − ∑ k = 1 ∞ 2 ( k + 2 ) 5 \displaystyle\sum_{k=1}^{\infty}\frac{k}{(k+2)^5}=\sum_{k=1}^{\infty}\frac{k+2}{(k+2)^5}-\sum_{k=1}^{\infty}\frac{2}{(k+2)^5}

= ∑ k = 1 ∞ 1 ( k + 2 ) 4 − 2 ∑ k = 1 ∞ 1 ( k + 2 ) 5 \displaystyle=\sum_{k=1}^{\infty}\frac{1}{(k+2)^4}-2\sum_{k=1}^{\infty}\frac{1}{(k+2)^5}

= ( π 4 90 − 1 − 1 16 ) − 2 ( ζ ( 5 ) − 1 − 1 32 ) \displaystyle=\left(\frac{\pi^4}{90}-1-\frac{1}{16}\right)-2\left(\zeta(5)-1-\frac{1}{32}\right)

= π 4 90 + 1 − 2 ζ ( 5 ) ≈ . 008 < . 01 \displaystyle=\frac{\pi^4}{90}+1-2\zeta(5)\approx.008\boxed{<.01} .

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@Maggie Miller Can u please elaborate which concept u used in the second step to arrive at pi^4/90-1-1/16

Tejas Suresh - 5 years, 10 months ago

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It's well-known that ∑ k = 1 ∞ 1 k 4 = π 4 90 \displaystyle\sum_{k=1}^{\infty}\frac{1}{k^4}=\frac{\pi^4}{90} (you can prove this with Fourier trig series, but I think for this problem it's ok to just take this value as fact).

Then

∑ k = 1 ∞ 1 ( k + 2 ) 4 = ∑ k = 3 ∞ 1 k 4 \displaystyle\sum_{k=1}^{\infty}\frac{1}{(k+2)^4}=\sum_{k=3}^{\infty}\frac{1}{k^4}

= ∑ k = 1 ∞ 1 k 4 − 1 1 4 − 1 2 4 \displaystyle=\sum_{k=1}^{\infty}\frac{1}{k^4}-\frac{1}{1^4}-\frac{1}{2^4}

= π 4 90 − 1 − 1 16 \displaystyle=\frac{\pi^4}{90}-1-\frac{1}{16} .

Maggie Miller - 5 years, 10 months ago

Did in jee style check for first 5 terms(without calculator :P ) you will get it

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