Sum of the series ∑r=1n rlog⁡r+1r is

Let tr=rlog⁡r+1r=r[log⁡(r+1)−log⁡r] =(r+1)log⁡(r+1)−rlog⁡r−log⁡(r+1)⇒ ∑r=1n tr=∑r=1n {(r+1)log⁡(r+1)−rlog⁡r}−∑r=1n log⁡(r+1) =(n+1)log⁡(n+1)−1log⁡1−log⁡(n+1)! =log⁡(n+1)n+1(n+1)!=log⁡(n+1)nn!

Sum of the series ∑r=1n rlog⁡r+1r is

Let tr=rlog⁡r+1r=r[log⁡(r+1)−log⁡r] =(r+1)log⁡(r+1)−rlog⁡r−log⁡(r+1)⇒ ∑r=1n tr=∑r=1n {(r+1)log⁡(r+1)−rlog⁡r}−∑r=1n log⁡(r+1) =(n+1)log⁡(n+1)−1log⁡1−log⁡(n+1)! =log⁡(n+1)n+1(n+1)!=log⁡(n+1)nn!

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Sum of the series $\sum_{r = 1}^{n} r \log \frac{r + 1}{r}$ is

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$\frac{\log (n + 1)}{n!}$

none of these

$n! \log (n + 1)$

$\log \frac{(n + 1)^{n}}{n!}$

answer is A.

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Detailed Solution

Let $t_{r} = r \log \frac{r + 1}{r} = r [\log (r + 1) - \log r]$
$\begin{array}{l} = (r + 1) \log (r + 1) - r \log r - \log (r + 1) \\ \Rightarrow \sum_{r = 1}^{n} t_{r} = \sum_{r = 1}^{n} {(r + 1) \log (r + 1) - r \log r} - \sum_{r = 1}^{n} \log (r + 1) \\ = (n + 1) \log (n + 1) - 1 \log 1 - \log (n + 1)! \\ = \log \frac{(n + 1)^{n + 1}}{(n + 1)!} = \log \frac{(n + 1)^{n}}{n!} \end{array}$