Cannot sum like this

Calculus Level 1

True or false ?

We have the following relations 1 + 2 + 3 + 4 + ⋯ + n = n 2 ( n + 1 ) 1 3 + 2 3 + 3 3 + ⋯ + n 3 = [ n 2 ( n + 1 ) ] 2 \begin{aligned} 1 +2+3+4+\cdots + n& = \dfrac{n}{2}\left(n+1\right)\\ 1^3+2^3+3^3+\cdots +n^3 & =\left[\dfrac{n}{2} \left(n+1\right)\right]^2 \end{aligned} If the infinite reciprocal sum of 1 + 1 1 + 2 + 1 1 + 2 + 3 + 1 1 + 2 + 3 + 4 + ⋯ = 2 . 1 +\dfrac{1}{1+2} +\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+\cdots =2\,. Then it must be true that 1 + 1 1 3 + 2 3 + 1 1 + 2 3 + 3 3 + 1 1 3 + 2 3 + 3 3 + 4 3 + ⋯ = 4 . 1+\dfrac{1}{1^3+2^3} +\dfrac{1}{1+2^3+3^3} +\dfrac{1}{1^3+2^3+3^3+4^3}+\cdots = 4 \,.

No Yes

Naren Bhandari by Naren Bhandari , 21, India

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1 solution

Jacopo Piccione
Oct 6, 2018

We don't have to solve the series: just comparing the terms we will know that we have to give N o \boxed{No} as answer.

Let a k a_k and b k b_k be the terms in position k k in first and second series. Then a k ≥ b k a_k\geq b_k , obviously because 1 a k ≤ 1 b k \frac{1}{a_k}\leq \frac{1}{b_k} . Since we're summing smaller terms, we can't end up with a bigger sum.

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