I know that $$\gamma=\lim_{n\to\infty }\left(\sum_{k=1}^n\frac{1}{k}-\ln(n)\right).$$ I'm trying to prove that we also have $$\gamma=\sum_{k=1}^\infty \left(\frac{1}{k}-\ln\left(1+\frac{1}{k}\right)\right).$$

So, $$\sum_{k=1}^n \left(\frac{1}{k}-\ln\left(1+\frac{1}{k}\right)\right)=\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^n\ln\left(1+\frac{1}{k}\right)=\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^n(\ln(k+1)-\ln(k)).$$ But $$\sum_{k=1}^n(\ln(k+1)-\ln(k))=\ln(n+1)$$ and thus, I get

$$\sum_{k=1}^\infty \left(\frac{1}{k}-\ln\left(1+\frac{1}{k}\right)\right)=\lim_{n\to\infty }\left(\sum_{k=1}^n\frac{1}{k}-\ln(n+1)\right),$$

hat's wrong here ?

  • $\begingroup$ The limits are the same - try subtracting one from the other. $\endgroup$ Sep 11, 2015 at 9:44
  • 1
    $\begingroup$ Let $H_n = \sum_{k = 1}^n \frac{1}{k}$. On the one hand, you know $\gamma = \lim\limits_{n\to\infty} H_n - \ln n$, on the other you get $\gamma = \lim\limits_{n\to\infty} H_n - \ln (n+1)$. But $\ln (n+1) - \ln n = \ln (1+1/n) \to 0$. $\endgroup$ Sep 11, 2015 at 10:10
  • $\begingroup$ Nothing's wrong, you just achieved the proof. $\endgroup$
    – user65203
    Sep 11, 2015 at 10:17
  • $\begingroup$ $\ln(n+1)=\ln n + O(1/n)$ as $n$ goes to infinity. $\endgroup$
    – Tom-Tom
    Sep 11, 2015 at 12:13

2 Answers 2


HINT: We have: $$\gamma=\lim_{n\to\infty}(\sum_{i=1}^{n}\dfrac{1}{i})-ln(n)$$ And: $$\gamma=\lim_{n\to\infty}\sum_{i=1}^{n}(\dfrac{1}{i}-ln(1+\dfrac{1}{i}))$$ Which is the same as: $$\gamma=\lim_{n\to\infty}\sum_{i=1}^{n}(\dfrac{1}{i})-\sum_{i=1}^{n}ln(1+\dfrac{1}{i})$$ We can subtract them from each other and get: $$\lim_{n\to\infty}ln(n)=\sum_{i=1}^{n}ln(1+\dfrac{1}{i})$$ Since $ln(a)+ln(b)=ln(ab)$: $$\lim_{n\to\infty}ln(n)=ln(\prod_{i=1}^{n}1+\dfrac{1}{i})$$ And thus: $$\lim_{n\to\infty}n=\prod_{i=1}^{n}1+\dfrac{1}{i}$$

  • $\begingroup$ I don't find this way of describing the solution good - you are starting by assuming what we have to prove, and you end up arriving at something that's clearly true. $\endgroup$
    – Wojowu
    Sep 11, 2015 at 11:53
  • $\begingroup$ Well, if it;s true then we can reason 'back' to the original equations, so... $\endgroup$
    – Mastrem
    Sep 11, 2015 at 11:57

Your problem just amounts to evaluating an easy telescoping product,



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