3
$\begingroup$

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 ?

$\endgroup$
4
  • $\begingroup$ 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

1
$\begingroup$

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: $$\lim_{n\to\infty}n=\prod_{i=1}^{n}1+\dfrac{1}{i}$$

$\endgroup$
2
  • $\begingroup$ 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
0
$\begingroup$

Your problem just amounts to evaluating an easy telescoping product,

$$\sum_{k=1}^n\ln\left(1+\frac1k\right)=\sum_{k=1}^n\ln\left(\frac{k+1}k\right)=\ln\left(\frac{n+1}{1}\right).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.