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# Interesting problem, need some help!

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

3 years, 10 months ago

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

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

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 :

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

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

= $$\boxed{ {(n + 1)}^2 f(n) - \frac{n(n + 1)}{2} }$$ · 3 years, 10 months ago

Isn't the series a diverging series ? · 3 years, 10 months ago