Can we pair all these fractions?

Algebra Level 3

Which of the following is larger?

$\begin{array} { c c c c c c c } &A =& \dfrac11 &+& \dfrac1{1+2} &+& \dfrac1{1+2+3} &+& \cdots \\ \phantom0 \\ &B =& \dfrac1{1^3} &+& \dfrac{1+2}{1^3+2^3} &+ &\dfrac{1+2+3}{1^3+2^3+3^3}& +&\cdots \\ \end{array}$

A

B

They are equal

1 solution

Naren Bhandari
May 8, 2018

$\small{B = \sum_{n=1}^{\infty}\left(\dfrac{ \frac{n(n+1)}{2}}{\frac{n^2(n+1)^2}{4}}\right) =\sum_{n=1}^{\infty} \dfrac{2}{n(n+1)} =\sum_{n=1}^{\infty}\dfrac{1}{\frac{n(n+1)}{2}} = A}$

Wonderful! Here's a harder question for you:

Prove that $\dfrac1{1^3} + \dfrac1{1^3 + 2^3} + \dfrac1{1^3 + 2^3 + 3^3} + \cdots = \dfrac43(\pi^2 - 9)$ .

Pi Han Goh - 3 years, 1 month ago

The sum can be written as $S =4 \sum_{n=1} ^{\infty} \dfrac{1}{n^2(n+1)^2} = 4\sum_{n=1}^{\infty} \underbrace{\left(\dfrac{1}{n^2} -2\left(\dfrac{1}{n} -\dfrac{1}{n+1}\right) + \dfrac{1}{(n+1)^2}\right)}_{\text{partial fraction decomposition}} =4 \left (\dfrac{\pi^2}{6} -2 + \dfrac{\pi^2}{6} -1\right) = 4\dfrac{2\pi^2 -18}{6} = \dfrac{4}{3}\,(\pi^2-9)$ By partial fraction .

Naren Bhandari - 3 years, 1 month ago

Nicely done again! Here's a harder one again:

Prove that $\dfrac1{1^2} + \dfrac1{1^2 + 2^2} + \dfrac1{1^2 + 2^2 + 3^2} + \cdots = 18 - 24 \ln 2$ .

Pi Han Goh - 3 years, 1 month ago

@Pi Han Goh – Actually, Sir this question I have calculated upto first 100 terms myself and I got about 1.36 like this and i think got something new and posted it here. However, Sir Mark Henning has provided so nice solution that amazed me.

Naren Bhandari - 3 years, 1 month ago

@Naren Bhandari – I don't understand, where's Mark Henning's solution?

Pi Han Goh - 3 years, 1 month ago

@Pi Han Goh – Reciprocal sum . Here is the problem.

Naren Bhandari - 3 years, 1 month ago

@Naren Bhandari – Oh wow. I didn't notice that.

Here's a following question if you're still interested (haha!):

Hence, prove that $e^{17} > 2^{24}$ and $\ln 2 > 0.7$ .

Pi Han Goh - 3 years, 1 month ago

@Pi Han Goh – I would like to try this too. However, I think should get some sleep. :) since I need to go for work.

Naren Bhandari - 3 years, 1 month ago

@Naren Bhandari – No problem man. Very clear work. Good night!

Pi Han Goh - 3 years, 1 month ago

@Pi Han Goh – Here is the problem with solution. Reciprocal sum . Moreover, I was amazed that its level 5 problem.

Naren Bhandari - 3 years, 1 month ago