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All values of $a_n = \log(\tfrac{n+1}{n})$ must be positive since $\tfrac{n+1}{n} > 1$. Hence $\sum_{n=1}^{\infty} a_n$ must be greater than $0$.

However when I calculate it as a telescoping sum, it seems to show that it equals $0$:

\begin{align*} \sum_{n = 1}^{\infty} \log(\tfrac{n+1}{n}) &= \sum_{n = 1}^{\infty} -\log(n) + \log(n+1)\\ &= -\log(1) + \log(2) - \log(2) + \log(3) - \log(3) + \log(4) ...\\ &= -\log(1)\\ &= 0 \end{align*}

So where am I going wrong?

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    $\begingroup$ Calculate a partial sum first. Does the resulting sequence converge? $\endgroup$ – David Mitra May 20 '16 at 11:32
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    $\begingroup$ You should write $=-\ln(1)+\ln(+\infty)=+\infty$. $\endgroup$ – user296113 May 20 '16 at 11:33
  • $\begingroup$ Calculate $S_N = \sum_{n = 1}^N \log(\tfrac{n+1}{n})$ first, then apply $\sum_{n = 1}^{\infty} \log(\tfrac{n+1}{n}) = \lim_{N\to \infty} S_N$. $\endgroup$ – CiaPan May 20 '16 at 11:49
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$$\sum_{n = 1}^{\infty} \log(\tfrac{n+1}{n}) = \sum_{n = 1}^{\infty} -\log(n) + \log(n+1) =\lim_{n\to\infty} ( -\log(1) + \log(2) - \log(2) + \log(3) - \log(3) + \log(4) ... = -\log(1) + ...+\log(n-1)-\log(n-1)+\log(n)) = \lim_{n\to\infty} \log(n)$$

There is always a last $\log(n)$ that is not being subtracted unless you add another $(-\log(n)+\log(n+1))$, but then $\log(n+1)$ remains.. So the final result shall be $$\sum_{n = 1}^{\infty} \log(\tfrac{n+1}{n})=\lim_{n\to\infty} \log(n) = \infty$$

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  • $\begingroup$ excellent thanks! $\endgroup$ – Brendan Hill May 20 '16 at 12:23
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For $x>-1$ we have $$\frac{x}{1+x}<\log(1+x)$$ Therefore$$\sum\frac{\frac1n}{1+\frac1n}<\sum \log\frac{n+1}{n}$$ Since $$\sum\frac{\frac1n}{1+\frac1n}=\sum\frac{1}{1+n}$$ diverges, your sum diverges too.

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The notation $\sum_{n=0}^{+\infty} a_n$ just means the limit $\lim_{k \to \infty} \sum_{n=0}^k a_n$ and it's a way to make sense of expressions like $a_0 + a_1 + a_2 + \dots$, which are formally meaningless. This is because a sum, and other operation as well, are only defined for a finite number of operand: $$a_0 + a_1 + a_2 := (a_0 + a_1) + a_2 = a_0 + (a_1 + a_2)$$

If you manipulate directly a summation with infinitely many terms you can end up with the most surprising and strange results. See this for example.

In general you can't easily calculate this limit but for a telescoping sum, as you know, the terms cancel out leaving a handy expression:

$$ \sum_{n=1}^{k} \log \frac{n+1}{n} = \sum_{n=1}^{k} [\log (n+1) - \log n ] = \log 2 - \log 1 + \log 3 - \log 2 + \\ +\dots + \log(k-1+1)-\log(k-1) + \log(k+1)-\log k = \log (k+1) $$

So the result is $$ \sum_{n=1}^{+\infty} \log \frac{n+1}{n} = \lim_{k \to +\infty} \sum_{n=1}^{k} \log \frac{n+1}{n} = \lim_{k \to \infty} \log (k+1) = +\infty $$

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