Interesting problem, need some help!

If $f(r) = 1 + \frac{1}{2} + \frac{1}{3} ....\frac{1}{r}$ , then what is the value of $\sum_{r=1}^{n} (2r+1)f(r)$

#NumberTheory #MathProblem #Math

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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}`	$\boxed{123}$

Comments

$\displaystyle \sum_{r = 1}^{n} (2r + 1)(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{r})$

= $\displaystyle \sum_{r = 1}^{n} \frac{2r + 1}{1} + \sum_{r = 2}^{n} \frac{2r + 1}{2} + \sum_{r = 3 }^{n} \frac{2r + 1}{3} + \dots + \sum_{r = (n - 1)}^{n} \frac{2r + 1}{(n - 1)} + \frac{2n + 1}{n}$

Consider :

$\displaystyle \sum_{r = i}^{n} (2r + 1)= \sum_{r = 0}^{n} (2r + 1) - \sum_{r = 0}^{i - 1} (2r + 1)$ $= {(n + 1)}^2 - i^2$

Our expression becomes :

$\displaystyle \sum_{i = 1}^{n} \sum_{r = i}^{n} \frac{2r + 1}{i} = \sum_{i = 1}^{n} \frac{{(n + 1)}^2 - i^2}{i}$

= $\displaystyle \sum_{i = 1}^{n} \frac{{(n + 1)}^2}{i} - \sum_{i = 1}^{n} i$

= $\boxed{ {(n + 1)}^2 f(n) - \frac{n(n + 1)}{2} }$

jatin yadav - 7 years, 8 months ago

Isn't the series a diverging series ?

Shivshankar D - 7 years, 8 months ago

It is, but the question asks for the partial sum.

gopinath no - 7 years, 8 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}`	$\boxed{123}$