# On computations around $\sum_{n=1}^N\frac{n\Lambda(n)}{n+N}$, where $\Lambda(n)$ is von Mangoldt function

By specialization with $F(x)=\frac{1}{1+x}$ in Apostol's Theorem 4.17 (Apostol, Introduction to Analytic Number Theory (Springer)), for intergers $N\geq 1$ one has $$\frac{\log N}{1+N}+\sum_{n=1}^N\frac{n\Lambda(n)}{n+N}=\sum_{n=1}^N\mu(n)G \left( \frac{N}{n} \right) ,$$ where $\Lambda(n)$ is von Mangoldt function, $\mu(n)$ is the Möbius function and $G(x)$ is defined by specialization in cited theorem as $$G(x):=\log x \left( \sum_{n\leq x}\frac{n}{n+x} \right).$$

Since $\lim_{N\to\infty}\frac{\log N}{N+1}=0$, I've asked to me about the convergence of the second term in LHS (or well, thus the term in RHS). After some computations with my computer, I see that is divergent, with a well defined slope, but I don't know how prove it, I say that I don't require help this time to get the average order (if you want/need compute it you are welcome to do it), it is enough this time, a method or computations to convince to me that is divergent.

On the other hand, then I've asked to me a second question, when I've divided the first equation by $N$ and after I've computed the limit as $N\to\infty$, that my computer say that $\frac{1}{N}\sum_{n=1}^N\frac{n\Lambda(n)}{n+N}$ converges to a positive contant. My computations were, since by definition we can deduce $\Lambda(even)\leq \log 2$ and $\Lambda(odd)\leq\log(odd)$ that $$\frac{1}{N}\sum_{n=1}^N\frac{n\Lambda(n)}{n+N}\leq\frac{1}{N} \left( \sum_{n=2k} \frac{n\log 2}{n+N} + \sum_{n=2k+1}\frac{n\log(2n+1)}{n+N}\right)$$ that is less than $$\frac{1}{N}\frac{ \left[ N \right] }{2}\frac{2}{2+N}+\frac{1}{N}\frac{ \left[ N \right] }{2}\frac{\log 3}{1+N}\to0,$$ as $N\to\infty$. Thus my computations seem wrongs about the upper limit of the previous series, I say the upper summation index.

Question. Can you give a reasoning that why $$\sum_{n=1}^N\frac{n\Lambda(n)}{n+N}$$ does diverge as $N\to\infty$? Can you give computations/reasonings to say if $$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N\frac{n\Lambda(n)}{n+N}$$ does equal to zero or a constant $c>0$? Thanks in advance.

• I don't know if such exercise was in the literature, please add a comment if it is. – user243301 May 9 '16 at 8:36

It can be easily solved using Abel's summation.We have $$\sum_{n\leq N}\frac{n\Lambda\left(n\right)}{n+N}=\frac{\psi\left(N\right)}{2}-N\int_{1}^{N}\frac{\psi\left(t\right)}{\left(t+N\right)^{2}}dt$$ where $\psi\left(x\right)=\sum_{n\leq x}\Lambda\left(n\right)$. Since from PNT we know that holds $\psi\left(x\right)\sim x$, we get $$\sim\frac{N}{2}-N\int_{1}^{N}\frac{t}{\left(t+N\right)^{2}}dt=\frac{N^{2}}{N+1}-N\log\left(2N\right)+N\log\left(N+1\right)$$ so $$\lim_{N\rightarrow\infty}\sum_{n\leq N}\frac{n\Lambda\left(n\right)}{n+N}=\infty$$ and $$\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{n\leq N}\frac{n\Lambda\left(n\right)}{n+N}=1-\log\left(2\right).$$
• Now I understand that it is possible take $f(x)=\frac{x}{x+N}$ in Abel's identity, I did not see this, now yes. Also I understand all details, onlyis summation and limits, the last was $1+\log\lim\frac{N+1}{2n}$. Very thanks much. – user243301 May 10 '16 at 15:06