How am I miscalculating the telescoping sum $\log(\frac{n+1}{n})$? 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?
 A: $$\sum_{n = 1}^{\infty} \log(\tfrac{n+1}{n})
 = \sum_{n = 1}^{\infty} -\log(n) + \log(n+1)
=\lim_{n\to\infty} ( -\log(1) + \log(2) - \log(2) + \log(3) - \log(3) + \log(4) ...
= -\log(1) + ...+\log(n-1)-\log(n-1)+\log(n))
= \lim_{n\to\infty} \log(n)$$
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 $$\sum_{n = 1}^{\infty} \log(\tfrac{n+1}{n})=\lim_{n\to\infty} \log(n) = \infty$$
A: 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.
A: 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
$$
\sum_{n=1}^{+\infty} \log \frac{n+1}{n} = \lim_{k \to +\infty} \sum_{n=1}^{k} \log \frac{n+1}{n} = \lim_{k \to \infty} \log (k+1) = +\infty
$$
