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

Note by Vikram Waradpande
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 :

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

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

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

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

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@Shivshankar D It is, but the question asks for the partial sum. Gopinath No · 3 years, 10 months ago

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