The converging series

Prove that: $\large \sum_{r=1}^{n}\frac{1}{r^2} < \frac{5}{3}$ and also Prove that $\large \sum_{r=1}^{n}\frac{1}{r^4} < \frac{10}{9}$ where $n$ is a positive integer.

#Algebra

Easy Math Editor

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Comments

For the first one, you can manipulate the series algebraically to create a new series for comparison, and then use telescoping sums to show that:

$\begin{aligned} \sum_{r=1}^n{\frac{1}{r^2}} & < 1 + \sum_{r=2}^n\underbrace{\left(\frac1{r - \frac{1}{2}} - \frac1{r + \frac{1}{2}}\right)}_{=\, \frac{1}{r^2 - \frac14} \,>\, \frac{1}{r^2}} \\ &= 1 + \frac23 - \frac1{n + \frac{1}{2}} < \frac53 \end{aligned}$

Zach Abueg - 3 years, 7 months ago

Thank you for writing but I am not able to understand your steps.Can you clearly explain your solution?

Vilakshan Gupta - 3 years, 7 months ago

Nice! I tried something similar for the other one, and it wasn't as pretty. Basically write (for $r \ge 2$ ): $\frac1{r^4} < \frac{16}{(4r^2-1)(4r^2-9)} = \frac1{3(2r-3)} - \frac1{2r-1} + \frac1{2r+1} - \frac1{3(2r-3)}$ and get $\sum_{r=1}^\infty \frac1{r^4} < 1 + \frac1{16} + \frac13 \left( \frac13 - \frac25 + \frac17 \right) = \frac{5483}{5040},$ which is good enough. Note that cutting it off one term earlier doesn't give a good enough estimate: $1 + \frac13\left( 1 - \frac23 + \frac15 \right) = \frac{10}9 + \frac1{15},$ alas.

Patrick Corn - 3 years, 7 months ago

@Patrick Corn – Sir, is this calculus? I actually don't know calculus. Is there any other method of solving it without calculus?

Vilakshan Gupta - 3 years, 7 months ago

I need help in doing these problems.It may be solved using calculus, but is there any other method which is without using calculus? @Chew-Seong Cheong @Anirudh Sreekumar @Brian Charlesworth @Zach Abueg Please help me out (anyone).

Vilakshan Gupta - 3 years, 7 months ago

@Zach Abueg , please explain your solution, i am unable to get it.

Vilakshan Gupta - 3 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$