Find the exact value of $n$ if $$\sum_{r=1}^{n}{\ln(\frac{r+1}{r})}=8.$$ So the sum is equal to $$\sum_{r=1}^{n}{\ln(1+\frac{1}{r})}=8.$$ $$\therefore\ln(1+1) + \ln(1+0.5)+...\ln(1+\frac1n)=8,$$ so $$(2)(1+0.5)...(1+\frac1n)=e^8$$ But I am unsure how then to solve the equation, and whether this is correct?
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1There does not exist such $n$ since $e^8$ is transcendental. – Oct 23 '18 at 15:27
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$$\sum_{k=1}^n \log\left({r + 1\over r}\right) = \sum_{k=1}^n \log(r + 1) - \log(r) = \ \log(n+1) - \log(1) = \log(n+1).$$
ncmathsadist
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HINT
Note that the summation telescopes and therefore
$$\ln\left(\frac{r+1}{r}\right)=\ln (r+1)-\ln r \implies \sum_{r=1}^{n}{\ln\left(\frac{r+1}{r}\right)}=\ln (n+1)$$
user
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Recall that $\ln a+\ln b = \ln(ab)$, so $$ \sum_{r=1}^n \ln\left(\frac{r+1}{r}\right) = \ln\left(\prod_{r=1}^n \frac{r+1}{r}\right) = \ln\left(\frac{2}{1}\frac{3}{2}\cdots\frac{n-1}{n}\frac{n+1}{n}\right) = \ln (n+1). $$ Can you finish?
EDIT: As pointed out in the comments, this shows that the original equation can be rewritten as $$ \ln(n+1) = 8, $$ which is not solved by any integer $n$.
MSDG
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Huh? Given that $n$ appears as an index in the sum it must be an integer. But there is no integer $n$ with $\ln(n+1)=8$. – David C. Ullrich Oct 23 '18 at 17:54
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@DavidC.Ullrich Oh that's a good point! I have not thought of that. That being said, the simplification shown in my post still holds. – MSDG Oct 23 '18 at 18:08
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Hint: Is there another way you can write $\ln \frac{a}{b}$?
Once you find it, try writing out a few terms of your sum.
John Hughes
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