Somemathhere

Prove that $$\sum_{i=1}^{n}\frac{(-1)^{i-1}}{i}{n\choose i}=H_n$$  where $H_n$ is the $n$-th harmonic number. Proof. Denotе the sum by $S$. Consider $$f(t)=\sum_{i=1}^{n}{n\choose i}t^{i-1}=\frac{(1+t)^n-1}{t}$$ Now integrate $$\int_{0}^{x} f(t)\mathrm{d}t=\sum_{i=1}^{n}{n\choose i}\frac{x^{i}}{i}=\int_{0}^{x}\frac{(1+t)^n-1}{t}\mathrm{d}t$$ Now plug $x=-1$ and we get $$\int_{0}^{-1}f(t)\mathrm{d}t=-S$$ On the other hand $$\int_{0}^{-1}f(t)\mathrm{d}t=\int_{0}^{-1}\frac{(1+t)^n-1}{t}\mathrm{d}t=-\int_{0}^{1}\frac{s^n-1}{s-1}\mathrm{d}s=\\=-\int_{0}^{1}1+s+s^2+\cdots+s^{n-1}\mathrm{d}s=-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)=-H_n$$

Let $f$ be a holomorphic function in $\{z\Big| 0<|z|<1\}$, continuous in $\{z\Big| 0<|z|\le 1\}$ and $|f(z)|\le \log\left(\frac{1}{|z|}\right)$. Prove that $f\equiv 0$. Proof. Since $f$ is continuous on the boundary, for $z_0$, s.t. $|z_0|=1$ we have $$|f(z_0)|=\lim_{z\to z_0}|f(z)|\le \lim_{z\to z_0}\log\left(\frac{1}{|z|}\right)=0$$ so $f$ is $0$ on the boundary. Consider now $g(z)=zf(z)$. So $g$ is also $0$ on the boundary. Now observe that $$\lim_{z\to 0}|g(z)|=\lim_{z\to 0}|z||f(z)|\le \lim_{z\to 0}|z|\log\left(\frac{1}{|z|}\right)=0$$ so $0$ is removable singularity for $g$. If we define $g(0)=0$, then $g$ is holomorphic on $\{z\Big||z|<1\}$, continuous on the boundary. From the maximum modulus principle $|g|$ attains its maximum at the boundary, and it is $0$ there, so $g\equiv 0$ on $\{z\Big||z|<1\}$. Now take $z_0\ne 0$, and then $f(z_0)=\frac{g(z_0)}{z_0}=0$, so $f\equiv 0$.