Let $f$ be a continuous, decreasing function, with $\displaystyle\lim_{x\rightarrow\infty}f(x)=0$.
Let $\gamma_k(x)=xf((k+1)x)-\int_{(k+1)x}^{(k+2)x}f(t)\mathrm{d}t,\displaystyle x>0$.
Let $\Gamma(x)=\sum\limits_{k=0}^{\infty}\gamma_k(x)$, suppose that $\displaystyle\lim_{x\rightarrow0} xf(x)=A$.
How does one show that $\displaystyle\lim_{x\rightarrow0}\Gamma(x)=A\gamma$ ? ($\gamma$ : euler constant)
This was in an exam I had earlier today, I can't figure out how to solve it.
Full exam questions here : http://mathb.in/20188
By scaling, we can assume that $A=1$. Let $\phi(x)=1/x$.
For any $\epsilon\gt0$, we can find an $x_\epsilon$ so that if $x\le x_\epsilon$, then $$ |xf(x)-1|\le\epsilon\tag{1} $$ Define $$ Sf(x)=\sum_{k=1}^\infty\left(xf(kx)-\int_{k}^{k+1}xf(tx)\,\mathrm{d}t\right)\tag{2} $$ Note that $$ xf(kx)-\int_{k}^{k+1}xf(tx)\,\mathrm{d}t=x\int_{k}^{k+1}[f(kx)-f(tx)]\,\mathrm{d}t\tag{3} $$ and since $0\le[f(kx)-f(tx)]\le[f(kx)-f((k+1)x)]$, we have $$ 0\le xf(kx)-\int_{k}^{k+1}xf(tx)\,\mathrm{d}t\le x[f(kx)-f((k+1)x)]\tag{4} $$ We sum $(4)$ for the appropriate $k$ to get $(5)$ and $(6)$.
For $n\le x_\epsilon/x$, we have $nx\,f(nx)\le1+\epsilon$ and therefore $$ \begin{align} 0\le\sum_{k=n}^\infty\left(xf(kx)-\int_{k}^{k+1}xf(tx)\,\mathrm{d}t\right)&\le xf(nx)\\ &\le\frac{1+\epsilon}{n}\tag{5} \end{align} $$ and $$ \begin{align} 0\le\sum_{k=n}^\infty\left(x\phi(kx)-\int_{k}^{k+1}x\phi(tx)\,\mathrm{d}t\right)&\le x\phi(nx)\\ &=\frac1n\tag{6} \end{align} $$ and since $|kx\,f(kx)-1|,|tx\,f(tx)-1|\le\epsilon$ for $k,t\le n$, we have $$ \begin{align} &\sum_{k=1}^{n-1}\left|\,\left[xf(kx)-x\phi(kx)\right]-\int_{k}^{k+1}\left[xf(tx)-x\phi(tx)\right]\,\mathrm{d}t\,\right|\\ &=\sum_{k=1}^{n-1}\left|\,\left[xf(kx)-\frac1k\right]-\int_{k}^{k+1}\left[xf(tx)-\frac1t\right]\,\mathrm{d}t\,\right|\\ &\le\sum_{k=1}^{n-1}\frac{2\epsilon}{k}\tag{7} \end{align} $$ For a given $\epsilon\gt0$ (since we are interested in making things small, assume $\epsilon\lt1$), let $x\le\epsilon x_\epsilon$, then inequalities $(5)$, $(6)$, and $(7)$ say that $$ \begin{align} |Sf(x)-S\phi(x)| &\le2\epsilon(\log(n)+1)+\frac{2+\epsilon}{n}\\ &\le2\epsilon\log(1/\epsilon)+5\epsilon\tag{8} \end{align} $$ Thus, by choosing $\epsilon\gt0$ small enough, we get $$ \lim_{x\to0}\left[Sf(x)-S\phi(x)\right]=0\tag{9} $$
Finally, for all $x\gt0$, we have $$ \begin{align} S\phi(x) &=\sum_{k=1}^\infty\left(x\phi(kx)-\int_{k}^{k+1}x\phi(tx)\,\mathrm{d}t\right)\\ &=\sum_{k=1}^\infty\left(\frac1k-\int_{k}^{k+1}\frac1t\,\mathrm{d}t\right)\\ &=\sum_{k=1}^\infty\left(\frac1k-\log\left(\frac{k+1}{k}\right)\right)\\ &=\gamma\tag{10} \end{align} $$ Therefore, combining $(9)$ and $(10)$, we get $$ \lim_{x\to0}Sf(x)=\gamma\tag{11} $$