Do you even telescope?

Algebra Level 2

Find the value of the following sum:

$\sqrt{\frac{1}{1^3}}+\sqrt{\frac{1}{1^3+2^3}}+\sqrt{\frac{1}{1^3+2^3+3^3}}+\ldots \infty$

Note that $\infty$ here refers to the series having infinite terms and not the value $\infty$ in the sum! In other words, the sum continues indefinitely.

Inspired by this problem.

The answer is 2.

1 solution

Prasun Biswas
Feb 1, 2015

We will use the following identity: $\displaystyle \sum_{i=1}^n i^3 = \left( \dfrac{n(n+1)}{2} \right)^2$

The given sum (say $S$ ) is equivalent to,

$S=\displaystyle \sum_{n=1}^\infty \left( \sqrt{\dfrac{1}{\displaystyle \sum_{i=1}^n i^3}} \right) = \displaystyle \sum_{n=1}^\infty \left( \sqrt{\left( \dfrac{2}{n(n+1)}\right)^2} \right) = \displaystyle \sum_{n=1}^\infty \left( \dfrac{2}{n(n+1)} \right) \\ \implies S=2\cdot \displaystyle \sum_{n=1}^\infty \left( \dfrac{1}{n(n+1)}\right)\\ \implies S=2\cdot\displaystyle \sum_{n=1}^\infty \left(\dfrac{1}{n} - \dfrac{1}{n+1} \right)$

This is indeed a telescoping sum as many of you would have already recognized and we can also see that only the first term of the first summation in $S$ would remain and all the other subsequent terms will get cancelled out. We get,

$S=2\times \frac{1}{1} = \boxed{2}$

It is technically wrong to separate the last sums with the telescoping terms because even if the difference converges, the sums individually diverge. Thus by separating them you are kind of writing $\infty - \infty$ .

Sudeep Salgia - 6 years, 4 months ago

Let's take an example. Consider the following sum:

$\displaystyle \sum_{i=1}^\infty \left( 1-1 \right)=\sum_{i=1}^\infty 1 + \sum_{i=1}^\infty 1$

Even though the individual summation parts diverge but when you consider the sum as a whole, the answer is $0$ .

P.S. - I had the same doubt a while ago, but you need to consider the sum as a whole and not as two different sums.

Prasun Biswas - 6 years, 4 months ago

That's precisely what I wanted to say. The expression makes sense only when written together and not as separate sums.I think it is better understood if written together making more sense as well. (Besides it saves on your latex. :D)

Sudeep Salgia - 6 years, 4 months ago

@Sudeep Salgia – Edited. Thanks. :)

Prasun Biswas - 6 years, 4 months ago

The last step should be done like this.

$S=2\left(\displaystyle \sum_{n=1}^{\infty} \left(\frac{1}{n}-\frac{1}{n+1}\right) \right)$

$S= 2\left(\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + .... + \frac {1}{\infty}\right)$

$S=2\times 1 = \boxed{2}$

Omkar Kulkarni - 6 years, 4 months ago

Yes, but I got bored of writing the $\LaTeX$ part for that and hence jumped right into the conclusion. That part is trivial though.

Prasun Biswas - 6 years, 4 months ago

Yeah, I meant that your method is wrong :P

Omkar Kulkarni - 6 years, 4 months ago

Do you even telescope?

The answer is 2.

1 solution

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