Show that the following sum is divergent $$\sum_{n=1}^{\infty}\ln\left(1+\frac1n\right)$$

I thought to do this using Taylor series using the fact that $$ \ln\left(1+\frac1n\right)=\frac1n+O\left(\frac1{n^2}\right) $$ Which then makes it clear that $$ \sum_{n=1}^{\infty}\ln\left(1+\frac1n\right)\sim \sum_{n=1}^{\infty}\frac1n\longrightarrow \infty $$ But I feel like I overcomplicated the problem and would be interested to see some other solutions. Also, would taylor series be the way you would see that this diverges if you were not told?

  • $\begingroup$ $\ln(1+1/n) \sim 1/n$ as $n \to +\infty$ means that for any $\epsilon > 0$ there is a $N$ such that for every $n > N$ : $\frac{1-\epsilon}{n} < \ln(1+1/n) < \frac{1+\epsilon}{n}$ $\endgroup$ – reuns Jul 30 '16 at 21:56
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    $\begingroup$ and no you can't write $\sum_n \ln(1+1/n) \sim \sum_n 1/n$ (it means nothing) but you can show that $\sum_{n < N} \ln(1+1/n) \sim \sum_{n < N} 1/n$ as $N \to +\infty$ $\endgroup$ – reuns Jul 30 '16 at 21:59
  • $\begingroup$ Why not? People seem to have understood $\endgroup$ – qbert Jul 30 '16 at 22:00
  • $\begingroup$ because $f(n) \sim g(n)$ as $n\to +\infty$ means something precise, namely that $\lim_{n \to \infty} \frac{f(n)}{g(n)} = 1$ $\endgroup$ – reuns Jul 30 '16 at 22:01
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    $\begingroup$ I really, really like this problem from a pedagogical perspective: as we can see below, it can be solved using such a wide variety of techniques in the standard precalculus / calculus toolbox, but it can't be solved by directly applying any of the convergence tests in the standard list given to students. In this way, I feel the sequences & series section of calc 2 is too algorithmic, with relatively little creativity required from students. This is the perfect sort of problem for cutting through that. Thanks for sharing @qbert $\endgroup$ – Kaj Hansen May 27 '19 at 5:58

Notice the following: $$\log\left(1+\frac{1}{n}\right)=\log\left(\frac{n+1}{n}\right)=\log(n+1)-\log(n)$$ Hence $$\sum_{k=1}^{n}\log\left(1+\frac{1}{k}\right)=\log(n+1) \to \infty$$

  • $\begingroup$ Duh ok thank you. Any response to my second question? $\endgroup$ – qbert Jul 30 '16 at 21:54
  • $\begingroup$ I think Taylor series would be the hint, for me at least. $\endgroup$ – preferred_anon Jul 30 '16 at 21:58
  • $\begingroup$ This is an efficient and compelling approach. +1 $\endgroup$ – Mark Viola Jul 30 '16 at 22:28

This is a special case of: Suppose $f(1)= 0, f'(1) > 0.$ Then $\sum f(1+1/n) = \infty.$

Proof: From the definition of the derivative (no Taylor necessary), we have

$$\frac{f(1+h)-f(1)}{h}= \frac{f(1+h)}{h} > \frac{f'(1)}{2}$$

for small $h>0.$ Thus

$$f(1+1/n) > \frac{f'(1)}{2}\cdot\frac{1}{n}$$ for large $n.$ By the comparison test we're done.

  • $\begingroup$ Nice generalization! $\endgroup$ – Kaj Hansen Jul 31 '16 at 1:06
  • $\begingroup$ Very cool. I think this is my favorite $\endgroup$ – qbert Jul 31 '16 at 1:32

Note that we have

$$\begin{align} \log\left(1+\frac1n\right)&=\int_n^{n+1}\frac{1}{t}\,dt\\\\ &\ge\frac1{n+1} \end{align}$$

and the harmonic series diverges.

But, suppose one forgoes that comparison and instead writes

$$\begin{align} \sum_{n=1}^{2^N-1}\int_n^{n+1} \frac{1}{t}\,dt&=\int_1^{2^N}\frac{1}{t}\,dt\\\\ &=\int_1^2 \frac{1}{t}\,dt+\int_{2}^{4}\frac{1}{t}\,dt+\dots+\int_{2^{N-1}}^{2^N}\frac{1}{t}\,dt\\\\ &\ge \frac12 (2-1)+\frac14 (4-2)+\dots +\frac{1}{2^N}(2^N-2^{N-1})\\\\ &=\frac{N}{2} \end{align}$$

which goes to $\infty$ as $N\to \infty$. And we are done! .

  • $\begingroup$ I really like using integrals to estimate series. Any books you'd recommend. I was only introduced to this method in complex variables $\endgroup$ – qbert Jul 30 '16 at 22:42
  • $\begingroup$ Pleased to hear that this was useful. I don't have a specific book to recommend. $\endgroup$ – Mark Viola Jul 30 '16 at 22:45
  • $\begingroup$ And I am trying to prove your first inequality. It seems clear that it is true for any positive increasing function in theintegrand, is that true? $\endgroup$ – qbert Jul 30 '16 at 22:46
  • $\begingroup$ Actually, the integrand is decreasing. Since $t\le n+1$, then $\frac1t \ge \frac{1}{n+1}$. Therefore, $$\int_n^{n+1}\frac{1}{t}\,dt\ge \frac{1}{n+1}\left((n+1)-n\right)$$ $\endgroup$ – Mark Viola Jul 30 '16 at 22:48

Applying a property of logarithms gives the equality $\displaystyle \sum_{n=1}^\infty \ln(1 + 1/n) = \ln \Bigg( \prod_{n=1}^\infty (1 + 1/n) \Bigg)$. Therefore, if $\displaystyle \sum_{n=1}^\infty \ln(1 + 1/n)$ converges, say to $c \in \mathbb{R}^+$, then $\displaystyle \prod_{n=1}^\infty (1 + 1/n)$ should converge to $e^c$.

Therein lies a contradiction: expanding this product yields a clearly divergent sum: the expansion will include a positive copy of $1/n$ for all $n \in \mathbb{N}$.

  • $\begingroup$ Super clever. Thank you! $\endgroup$ – qbert Jul 30 '16 at 21:59
  • $\begingroup$ Glad I could help! $\endgroup$ – Kaj Hansen Jul 30 '16 at 21:59

$$\sum_{n=1}^{m}\log\left(\frac{n+1}{n}\right)=\sum_{n=1}^{m}(\log(n+1)-\log n)=\log(m+1)$$ The partial sums clearly diverge. Alternatively using the cauchy condensation test the series converges iff the series$$\sum_{n=1}^{\infty}2^{n}\ln(1+1/2^{n})$$ converges. The transformed series diverges since the terms don't go to zero and so the original series diverges.

  • $\begingroup$ +1. It's the straightforward way to deal with the question. $\endgroup$ – Felix Marin Aug 1 '16 at 4:11

In THIS ANSWER, I showed using elementary analysis only that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\le\log(x)\le x-1\tag1$$

Letting $x=1+\frac1n$ in $(1)$ reveals

$$\frac1{n+1}\le\log\left(1+\frac1n\right)\le \frac1n\tag2$$

And using the left-hand side of $(2)$ shows that

$$\sum_{n=1}^N \log\left(1+\frac1n\right)\ge \sum_{n=1}^N \frac1{n+1}$$

Inasmuch as the harmonic series diverges, we see that the series of interest diverges also. And we are done!


"Sophisticated" does not mean "complicated". In my opinion, despite using more sophisticated ideas (asymptotic analysis), your proof is simpler than all of the other current answers — even the one expressing it as a telescoping series.

Incidentally, you possibly made an oversight: to complete the proof,

$$ \sum_{n=1}^{\infty} O\left(\frac{1}{n^2} \right) = O\left( \sum_{n=1}^{\infty} \frac{1}{n^2} \right) = O(1)$$

(also, a remark: for this argument to be valid, it's important that the $O$ on the left is uniform; e.g. the same 'hidden constant' works for all $n$)

  • $\begingroup$ Sorry I'm not sure I follow/may have used notation incorrectly. Are you saying that the final identity is ok since $(1/n^p)/(1/n^2)=c$ or that we even need to worry about how this c varies with p? $\endgroup$ – qbert Jul 31 '16 at 1:35
  • $\begingroup$ @qbert: The final identity holds because the sum is convergent. It's a constant! $\endgroup$ – user14972 Jul 31 '16 at 1:42
  • $\begingroup$ Ok, got it. Thanks! Also, what subjects use big o frequently? $\endgroup$ – qbert Jul 31 '16 at 1:43
  • $\begingroup$ @qbert: For people who use the notation, it can come up in just about any field of mathematics. Often, there are many different ways to say the same thing (albeit with varying amounts of facility), and people will use the ones they're most comfortable with. However, off the top of my head, "analysis of algorithms" is the only subject I can think of where practically everybody uses it. $\endgroup$ – user14972 Jul 31 '16 at 1:47
  • $\begingroup$ @qbert: For example, in $5$-adic analysis, $3$ and $53$ are 'nearby' integers. Some people like to emphasize the arithmetic statement: $3 \equiv 53 \pmod{5^2}$. Some people like to measure the distance: $d(3, 53) = 5^{-2}$. Others prefer to express the that they are the same up to some amount of precision: $3 = 53 + O(5^2)$. And each of these ways of thinking themselves have several ways they could be notated. $\endgroup$ – user14972 Jul 31 '16 at 1:53

You can also use the Abel's summation $$\sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)=\sum_{n=1}^{N}1\cdot\log\left(1+\frac{1}{n}\right)=N\log\left(1+\frac{1}{N}\right)+\int_{1}^{N}\frac{\left\lfloor t\right\rfloor }{t\left(t+1\right)}dt $$ where $\left\lfloor t\right\rfloor $ is the floor function. Since $\left\lfloor t\right\rfloor =t+O\left(1\right) $ we have $$\sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)=N\log\left(1+\frac{1}{N}\right)+\log\left(N+1\right)+O\left(1\right) $$ and taking $N\rightarrow\infty$ we can see that the series diverges.


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