Problem  Prove that $$ \sum_{p \leq x} {\log p \over p} = \log x + O(1) $$ as $x \to \infty$.  Notes: $p$ ranges over primes, $\log$ is natural  Progress  Using Riemann-Stieltjes integration and integration by parts, I think you can write the sum as $$ \int_2^x {\log t \over t} d\pi(t) = O(1) - \int_2^x {1 - \log t \over t^2} \pi(t) dt. $$ From here, I tried using the prime number theorem, specifically $$ \pi(x) = \operatorname{li}(x) + O(xe^{-c\sqrt{\log x}}), $$ together with $$ \operatorname{li}(x) = {x \over \log x} + {x \over \log^2 x} + O\Big({x \over \log^3 x}\Big). $$ All the terms cancel out perfectly except for the integrals of the error terms from the prime number theorem, which diverge. If they converged, they could go into the $O(1)$ term, which would finish the proof.  Please note that this is tagged as homework.    Those integrals actually do converge. In fact, $$ O(xe^{-c\sqrt{\log x}}) = O\Big({x \over \log^3 x}\Big), $$ so there is no need to even consider them separately. I will submit this as an answer to the question this is a duplicate of now.
120551
I saw reference to this result of Chebyshev's: $$\sum_{p\leq n} \frac{\log p}{p} \sim \log n \text{ as }n \to \infty,$$ and its relation to the Prime Number Theorem. I'm looking into  I was wondering if anyone could give me a sense for how difficult or involved the usual proof is.  I don't really need to see the whole thing in detail, more wondering about the general difficulty/complexity of the result. Thanks!
30361
How do I determine this limit&nbsp;?:  $$ \lim_{x \to \infty}\left[x - x^{2}\log\left(1 + {1 \over x}\right)\right] $$  I have tried to decompose the $\log$ function but I don't know how to proceed from there.
68886
$\sum_{n=1}^\infty \log \frac{n+1}{n}$  $\sum_{n=1}^\infty \log \frac{n+1}{n}$ =  $\displaystyle\lim_{n \to{+}\infty}(\log2 - \log1)+(\log3-\log2)+...+(\log(n+1)-\log n)$=$\displaystyle\lim_{n \to{+}\infty}\log(n+1)\to \infty$. So, the series diverges.  Is my procedure correct?
39730
This integral below $$ I:=\int_0^\infty \log^2 x\frac{1+x^2}{1+x^4}dx=\frac{3 \pi^3}{16 \sqrt 2} $$ is what I am trying to prove.  Thanks.  We can not expand the denominator as a series since the domain of integration is for $x\in [0,\infty)$.  Next I wrote $$ I=\int_0^\infty \log^2 x \frac{1+x^4-x^4+x^2}{1+x^4}dx=\int_0^\infty \log^2x \left(\frac{1+x^4}{1+x^4}+\frac{x^2-x^4}{1+x^4}\right)dx=\\ \int_0^\infty \log^2 x \, dx+\int_0^\infty \log^2 x \frac{x^2}{1+x^4}dx-\int_0^\infty \log^2 x \frac{x^4}{1+x^4}dx, $$ however only the middle integral is convergent.  I am not sure how to go about solving this problem.  Thank you
15598
Classify  $$\sum_{n=3}^\infty \frac{1}{n\log(n)}$$ as absolutely convergent, conditionally convergent or divergent.  Is it, $$\sum_{n=3}^\infty \frac{1}n$$ is a divergent $p$-series as $p=1$, and $$\lim_{n\to\infty} \frac{1}{n\log(n)}{n} = 0$$ by comparison test. And this converges to $0$.  So, $$\sum_{n=3}^\infty \frac{1}{n\log (n)}$$ is conditionally convergence?  I'm not sure if I'm doing right or not. Could you guide me?  Thanks in advance! :)
104451
The problem I'm trying to solve is:  Let $1\leq p, q\leq\infty$ such that $\frac{1}{q}+\frac{1}{p}=1$. If  $$\sum_{n=1}^{\infty} |x_n||y_n|&lt;\infty$$ for all $(x_n)\in l^p$, then $(y_n)\in l^q$.  $\bullet$ If $p=\infty$, then picking $(x_n)=(1,1,1,...)\in l^\infty$, we have $$\sum_{n=1}^{\infty} |y_n|&lt;\infty$$ i.e. $(y_n)\in l^1$.  but how can I prove for $1\leq p&lt;\infty$?
97748
Let $$A(p,q) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q},$$ where $H^{(p)}_n = \sum_{i=1}^n i^{-p}$, the $n$th $p$-harmonic number.  The $A(p,q)$'s are known as alternating .     Can someone provide a nice proof that    $$A(1,1) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2?$$   I worked for a while on this today but was unsuccessful.  Summation by parts, swapping the order of summation, and approximating $H_k$ by $\log k$ were my best ideas, but I could not get any of them to work.  (Perhaps someone else can?)  I would like a nice proof in order to complete .  Bonus points for proving $A(1,2) = \frac{5}{8} \zeta(3)$ and $A(2,1) =  \zeta(3) - \frac{1}{2}\zeta(2) \log 2$, as those are the other two alternating Euler sums needed to complete my answer.   Added: I'm going to change the accepted answer to robjohn's $A(1,1)$ calculation as a proxy for the three answers he gave here.  Notwithstanding the other great answers (especially the currently most-upvoted one, the one I first accepted), robjohn's approach is the one I was originally trying.  I am pleased to see that it can be used to do the $A(1,1)$, $A(1,2)$, and $A(2,1)$ derivations.
22012
$${\pi \over 2} \le \sum_{n=0}^{\infty} {1 \over {1+n^2}} \le {\pi \over 2} + 1$$  I see I should use Riemann sum, and that $$\int_0^{\infty} {dx \over {1+x^2}} \le \sum_{n=0}^{\infty} {1 \over {1+n^2}}$$  But how do I explain this exactly, and how to get the other half of the inequality (with ${\pi \over 2} + 1$)?
70707
A while ago, I started to look at expressions of the following form: $$ S_p:=\sum_{n=1}^\infty \frac1n - \frac1{n+1/p}, $$  where $p$ is prime, because otherwise things get too complicated for me at the moment.  What I found so far is the following: $$ S_p= p -\log(2p) - \frac12 \pi \cot\left(\frac\pi p\right) + T_p, $$ where $T_p$ contains $\frac{p-1}2$ terms $t_{p,k}$ of the form:  \begin{cases} \pm 2 \sin\left(\frac{k\pi}{2p}\right) \log\left(\sin\left(\frac{k\pi}{2p}\right)\right) \\ \pm 2 \cos\left(\frac{k\pi}{2p}\right) \log\left(\sin\left(\frac{k\pi}{2p}\right)\right)\\ \pm 2 \sin\left(\frac{k\pi}{2p}\right) \log\left(\cos\left(\frac{k\pi}{2p}\right)\right)  \\ \pm 2 \cos\left(\frac{k\pi}{2p}\right) \log\left(\cos\left(\frac{k\pi}{2p}\right)\right)  \end{cases}   An example can be found .     How does the closed form for $\sum_{n=1}^\infty \frac1n - \frac1{a+n}$ look like? (EDIT: where $a=1/p$)
32183
How can we prove that:     $$\int_0^1 \frac{\log(x)\log(1-x)\log^2(1+x)}{x}dx=\frac{7\pi^2}{48}\zeta(3)-\frac{25}{16}\zeta(5)$$ where $\zeta(z)$ is the .   The best I could do was to express it in terms of Euler Sums. Let $I$ denote the integral.  $$I=-\frac{\pi^2}{24}\zeta(3)+2\sum_{r=2}^\infty \frac{(-1)^r (H_r)^2}{r^3}-2\sum_{r=2}^\infty \frac{(-1)^r H_r}{r^4}+2 \sum_{r=2}^\infty \frac{(-1)^r H_r^{(2)}H_r}{r^2}-2\sum_{r=2}^\infty \frac{(-1)^r H_r^{(2)}}{r^3}$$  where $\displaystyle H_r^{(n)}=\sum_{n=1}^r \frac{1}{k^n}$. I am unable to simplify these sums further. Does anyone have any idea on how to solve this integral?  Please see  for more details.    Some time ago I was able to solve the simpler integral:     \begin{align*} \int_0^1 \frac{\log(1-x)\log(x)\log(1+x)}{x}dx &amp;=-\frac{3 \pi^4}{160}+\frac{7\log(2)}{4}\zeta(3)-\frac{\pi^2 \log^2(2)}{12} +\frac{\log^4(2)}{12} \\ &amp;\quad+ 2 \text{Li}_4 \left(\frac{1}{2} \right) \end{align*} where $\text{Li}_n(z)$ is the .
51806
Hi I am trying to prove$$ I:=\int_0^1 \log\left(\,\Gamma\left(x+\alpha\right)\,\right)\,{\rm d}x =\frac{\log\left(2\pi\right)}{2}+\alpha \log\left(\alpha\right) -\alpha\,,\qquad \alpha \geq 0. $$ I am not sure whether to use an integral representation or to somehow use the Euler reflection formula $$ \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin \pi z} $$ since a previous post used  that to solve  this kind of integral.  Other than this method, we can use the integral representation  $$ \Gamma(t)=\int_0^\infty x^{t-1} e^{-x}\, dx. $$  Also note $\Gamma(n)=(n-1)!$.
104853
I would like to prove that $\sum_{k=1} ^ n k^{-1} = \ln(n) + O(1)$. That is, I would like to show that there is some natural number $N$ large enough so that $n \ge N$ implies:  $$|\sum_{k=1}^n k^{-1} - \ln(n) | &lt; M,$$  where $M$ is a positive constant independent of $n$. I think the idea behind the argument must be easy, but I am somehow missing it. I do see that $\sum_{k=1}^n k^{-1}$ can be taken as a sort of approximation, using areas of rectangles, to $\int_1^n \frac{1}{x}dx = \ln(n)$. So maybe the argument uses the integral comparison test? But, anyway, I'm still fumbling with how to proceed.  Hints or solutions are greatly appreciated.
89219
Could someone provide, please, a proof of the theorem below?  "Being $x$ and $b$ integers greater than $1$, which can not be represented as powers of the same basis (positive integer) and integer exponent, then the logarithm of $x$, in base $b$, is an irrational number."  Well... Assuming that the logarithm is a rational number $p/q$ ($p$ and $q$ are relatively prime integers), I know I can write $x^q = b^p$, but I can not conclude from this fact that $b$ and $x$ are powers of a same integer (and with integer exponent).
11505
Find the values of $p$ for which the series $\sum_{n=2}^\infty \frac{1}{n(\ln n)^p}$ is convergent. I know that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent if $p&gt;1$ and divergent if $p\leq1$. So with the series in question, I can't just say &quot;when $p&gt;1$&quot; right? Because the denominator is $n(\ln n)^p$ not just $n^p$ as the formula states. So how should I go about tackling this problem? I know that doing a u-sub for $\ln n$ would work because $du$ would equals $\frac{dn}{n}$ but what would I do with the $p$? Thanks for the help.
17255
Is $\sum_\limits{p}^{\infty}\frac{1}{p^{2}}$ irrational where p is prime? How to prove it?
106773
Does $\sum\limits_{n=3}^{\infty} \dfrac{1}{(\log\log n)^{\log n}}$ converge or diverge ? I tried some tests , but nothing conclusive is coming . Pleas help
56035
Question  Let $X_n\geq 0$ be independent for $n\geq 1$. The following are equivalent.  $\sum_{n=1}^\infty X_n&lt;\infty$ a.s. $\sum_{n=1}^\infty[P(X_n&gt;1)+E(X_nI(X_n\leq 1)]&lt;\infty$ $\sum_{n=1}^\infty E(X_n/(1+X_n))&lt;\infty$   My attempt I was able to prove $(1)$ implies $(2)$ and $(2)$ implies $(3)$ but not $(3)$ implies $(1)$. For $(1)$ implies $(2)$, observe that by Borel-Cantelli $\sum_{n=1}^\infty P(X_n&gt;1)&lt;\infty$, else $X_n&gt;1$ infinitely often w.p.1 so that $(1)$ is violated. To see that $\sum_{n=1}^\infty E(X_nI(X_n\leq 1)]&lt;\infty$, apply Kolmogorov three-series theorem (I think). For $(2)$ implies $(3)$ we can write that $$ \begin{align} \frac{X_n}{1+X_n}&amp; =\frac{X_n}{1+X_n}I(X_n\leq 1)+\frac{X_n}{1+X_n}I(X_n&gt;1)\\ &amp;\leq X_nI(X_n\leq 1)+ I(X_n&gt;1). \end{align} $$ Now take the expectation of both sides and sum on $n$ using $(2)$ to conclude $(3)$. For the last implication $(3)$ implies $(1)$, I tried to use Kolmogorov three series theorem. To this end put $Y_n=X_nI(X_n\leq 1)$. First note that $\sum_{n=1}^\infty P(X_n&gt;1)&lt;\infty$ since $$ P(X_n&gt;1)\leq P\left(\frac{X_n}{1+X_n}&gt;\frac{1}{2}\right)\leq 2E\frac{X_n}{1+X_n} $$ by Markov's inequality. But I am unable to show that $\sum EY_n&lt;\infty$ and $\sum\text{Var}(Y_n)&lt;\infty$. Any help is appreciated.