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I'm trying to figure out whether

$$\sum_{x = 1}^{\infty}\ln\bigg(\frac{x}{x+1}\bigg)$$

converges or diverges.

Here is what I tried doing:

\begin{align}\sum_{x = 1}^{\infty}\ln\bigg(\frac{x}{x+1}\bigg) &= \sum_{x = 1}^{\infty}\ln(x)-\ln(x+1) \\ &= \sum_{x = 1}^{\infty}\ln(x) - \sum_{x = 1}^{\infty}\ln(x+1) \\ &=(\ln(1) + \ln(2) + \cdots)-(\ln(2)+\ln(3) + \cdots) \\ &=\ln(1)\end{align}

So I though that the series would converge to $\ln(1) = 0$ since all of the other terms cancel out. I checked using WolframAlpha and the series actually diverges. I was wondering why is this the case and what is wrong with my work above?

Xander Henderson
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user262291
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    Hint: Use the telescoping idea to calculate a generic partial sum. See what's the limit of those partial sums. – Jyrki Lahtonen Mar 19 '18 at 17:57
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    You can't permute terms in a conditionally convergent series. – Bernard Mar 19 '18 at 17:58
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    And, if the penny didn't drop, consider the following telescoping sum $$(1^2-0^2)+(2^2-1^2)+(3^2-2^2)+\cdots=\sum_{n=0}^\infty(2n+1).$$ Does that converge? – Jyrki Lahtonen Mar 19 '18 at 18:00
  • I can see why it diverges right now. So is the problem with what I did above is rearrange the terms of a conditionally convergent series? – user262291 Mar 19 '18 at 18:02
  • Hint: All of the terms are negative so you can safely rule out your results. – Michael Paris Mar 19 '18 at 18:03
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    The way I look at it you misused the following result. The theorem $$\sum_n(a_n+b_n)=\sum_n a_n+\sum_n b_n$$ only holds if at least two of those series converge (when the theorem implies that the third one converges also). Conditional convergence didn't enter the argument here IMHO. You used this result on the fifth line of your post. But you haven't established the convergence of any of those series (they all diverge). – Jyrki Lahtonen Mar 19 '18 at 18:06
  • OK, now I see how that might also be interpreted as a reordering of a conditionally convergent series. I take that back. – Jyrki Lahtonen Mar 19 '18 at 18:09

2 Answers2

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Your attempt is incorrect at this step: $$ \sum_{x = 1}^{\infty}\ln(x)-\ln(x+1) \color{red}{\overset{?}{=}} \sum_{x = 1}^{\infty}\ln(x) - \sum_{x = 1}^{\infty}\ln(x+1) $$ In this step, you have rearranged the terms of the series in order to get two new series. This is not generally possible—see, for example, the Riemann rearrangement theorem. You make another mistake when you write $$ \sum_{x = 1}^{\infty}\ln(x) - \sum_{x = 1}^{\infty}\ln(x+1) = (\ln(1) + \ln(2) + \cdots)-(\ln(2)+\ln(3) + \cdots) \color{red}{\overset{?}{=}} \ln(1). $$ This only makes sense if the two series both converge. Unfortunately, neither series converges (the series $ \sum_{n=1}^{\infty} \ln(n)$ diverges to infinity), and your sum-of-series looks like $$ \sum_{x = 1}^{\infty}\ln(x) - \sum_{x = 1}^{\infty}\ln(x+1) "=" \infty-\infty, $$ which is an indeterminate form. That is, the expression $\infty-\infty$ is not well-defined, and can take on any value, depending on the underlying construction.

Recall the definition of a convergent series:

We say that $$ \sum_{n=1}^{\infty} a_n = S $$ for some finite $S$ (that is, the series converges to a finite value $S$) if and only if the sequence of partial sums converges to $S$. That is, with $$ S_N := \sum_{n=1}^{N} a_n, $$ we have $\lim_{N\to\infty} S_N = S$.

We should therefore reason as follows: \begin{align} \sum_{x=1}^{\infty} \ln(x)-\ln(x+1) &= \lim_{N\to\infty} \sum_{x=1}^{N} \ln(x) - \ln(x+1) \\ &= \lim_{N\to\infty} \left[ (\ln(1) - \ln(2)) + (\ln(2)-\ln(3)) + \dotsb (\ln(N) - \ln(N+1)) \right] \\ &= \lim_{N\to\infty} \left[ \ln(1) + (-\ln(2) + \ln(2) + \dotsb (-\ln(N) + \ln(N)) - \ln(N+1) \right] \\ &= \lim_{N\to\infty} \left[ 0 + 0 + \dotsb + 0 - \ln(N+1) \right] \\ &= \lim_{N\to\infty} \left[ -\ln(N+1) \right] \\ &= -\infty. \end{align} The important point here is that we are carefully examining the series as a limit of partial sums. As we do this, we are only using the associativity of addition to get the result, then taking a limit after we have determined the general form of the $N$-th partial sum. When we do this, we don't lose the "extra" $\log(N+1)$ at the end of the series, which is lost in your formulation.

Xander Henderson
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    +1 for pointing out the spot where the OP erred. I was apparently typing a comment with similar content (but less detail) while you were typing your post. – Jyrki Lahtonen Mar 19 '18 at 18:11
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Note that

$$\sum_{x=1}^X \log(x)-\log(x+1)=\log(1)-\log(X+1)$$

The logarithmic divergence should not be surprising inasmuch as

$$\log\left(\frac{x}{x+1}\right)=-\log\left(1+\frac1x\right)\le -\frac{1}{x+1}$$

and the harmonic series diverges logarithmically.

Mark Viola
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