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First of all, the Taylor expansion for $\ln(1+x)$ is

$$\ln(1+x)=\sum _{k=1}^{\infty } \frac{(-1)^{k+1} x^k}{k}$$

It follows that

$$\lim_{x\to-1}\left(\sum _{k=1}^{\infty } \frac{(-1)^{k+1} x^k}{k}-\ln(1+x)\right)=0$$

consequently,

$$\lim_{t\to1}\left(\sum _{k=1}^{\infty } \frac{ t^k}{k}-\ln (1-t)\right)=0$$

$$\lim_{t\to\infty}\left(\sum _{k=1}^{t } \frac{ 1}{k}+\ln t\right)=0$$

On the other hand, we have

$$\lim _{{t\rightarrow \infty }}\left(\sum _{{k=1}}^{t}{\frac {1}{k}}-\ln t\right)=\gamma$$

So where is the error?

Anixx
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  • The error is the "It follows" statement. – zhw. Apr 04 '15 at 03:30
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    How do you know that the limit is $0$? Be cautious about where the sum converges. – Michael Burr Apr 04 '15 at 03:31
  • How does the first consequently statement imply the second? – Teoc Apr 04 '15 at 03:33
  • Also note that $$\sum {k=1}^{\infty } \frac{ t^k}{k}=-\ln (1-t)$$ so $$\sum _{k=1}^{\infty } \frac{ t^k}{k}-\ln (1-t)=-2\ln (1-t)$$ and it is clearly incorrect to have $$\lim{t \to 1}-2\ln (1-t)=0$$ Might also be worth pointing out that the limit you are taking does not exist from the left – graydad Apr 04 '15 at 03:42
  • Interchanging limits (sums, integrals, limits, etc.) always needs justification. – Batman Apr 04 '15 at 03:44
  • You taking an analytic function, $f(x)=\log(1+x)$, and assuming that its power series in $x=0$ converges to $f(x)$ for $x=-1$, having an absolute value that equals the radius of convergence. That cannot hold, since $\log(1+x)$ has a wild singularity in $x=-1$. – Jack D'Aurizio Apr 04 '15 at 03:53

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