All Questions
Tagged with logarithms summation
244
questions
3
votes
3
answers
385
views
$\operatorname{Li}_{2} \left(\frac{1}{e^{\pi}} \right)$ as a limit of a sum
Working on the same lines as
This/This and
This
I got the following expression for the Dilogarithm $\operatorname{Li}_{2} \left(\frac{1}{e^{\pi}} \right)$:
$$\operatorname{Li}_{2} \left(\frac{1}{e^{\...
- 862
0
votes
1
answer
33
views
Natural Log's Property Doesn't Transfer Over
I am trying to rewrite the summation of $\ln(x)$ equation into a continuous function using logarithmic properties. We already know that $\left(\sum_{n=1}^{x}\ln\left(n\right)\right)$ is just equal to $...
- 1
2
votes
2
answers
144
views
Is there a closed form solution to $\sum_{n=1}^{\infty}\frac{\ln\left(\frac{n+1}{n}\right)}{n}$ [duplicate]
While coming up with an idea for another way to milk the integral in my previous question, I got stuck at this summation: $$\sum_{n=1}^{\infty}\frac{\ln\left(\frac{n+1}{n}\right)}{n}$$
I do not know ...
- 1,688
0
votes
1
answer
61
views
What property is being used to simplify this logarithm? $ 4^{log_2 n} = n^2$
I had this equation for calculating this recursion:
$$ T(n) = 4T(\frac{n}{2}) + n $$
I'm using a recursion tree to solve this, it's similar to this one:
But anyway what matters is that from the ...
- 33
0
votes
0
answers
43
views
Compute a tight upper bound of $\sum_{i=1}^{n-1}\frac{1}{3^i\log{n}- 3i}$?
I am trying to compute a tight upper bound of the sum below.
$\sum_{i=1}^{n-1}n\frac{\frac{1}{3^i}}{\log_3{(n/3^i)}}$
I was able to 'simplify' it up to the expression below.
$n\sum_{i=1}^{n-1}\frac{1}{...
- 185
1
vote
2
answers
72
views
Closed form of sum with ceil of logarithm.
I've been practicing discrete math recently and I'm stuck on this problem. Could someone help me with this, give me some hint or direction? The problem is:
Find the closed form of $\sum_{k=1}^n \lceil{...
- 145
0
votes
0
answers
79
views
What is the name for a matrix which is generated by a recursive sum whose form equals a recursive product when replacing the sums with products?
In this answer to the question "Do these series converge to logarithms?" it is shown by George Lowther that each Dirichlet series involving the pattern of divisors converge to $\log(n)$ in ...
- 7,420
0
votes
2
answers
160
views
$\lim_{{n \to \infty}} \sum_{{k=1}}^{n} \arctan\left(\frac{1}{k}\right) - \ln n$
$\arctan(x) = \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1}x^{2m+1}$
\begin{align*}
\arctan\left(\frac{1}{k}\right) &= \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1}\left(\frac{1}{k^{2m+1}}\right)
&= \frac{...
0
votes
0
answers
19
views
Correct approach on log(n+m+z)!
So I have for simpler notation, for a fixed infinitely large natural number $n$ and all finite natural numbers $m,z$, then $L(n+m)=\log(n+m)!$, and this is equal to $$L(n+m)=\log n!+\sum_{k=1}^m\log(n+...
- 243
4
votes
1
answer
123
views
Asymptotic formula/closed form of $\sum_{r=0}^{n}\frac{(-1)^r\binom{n}{r}\log(n+r+1) }{n+r+1}$
I need an asymptotic formula/closed form for the sum $$\sum_{r=0}^{n}\frac{(-1)^r\binom{n}{r}\log(n+r+1) }{n+r+1}$$ where $n\in\mathbb{N}$
Denote $$S_n=\sum_{r=0}^{n}\frac{(-1)^r\binom{n}{r}\log(n+r+1)...
- 926
4
votes
1
answer
175
views
How can I evaluate this infinite sum?
I was tackling an integral I saw on the Maths 505 YouTube channel and I came across this sum.
$$\sum_{n=1}^\infty(\frac{1}{2n}+1-(n+1)\ln(1+\frac{1}{n}))$$
I plugged it into Wolfram Alpha and it gave ...
- 55
0
votes
1
answer
48
views
How to simplify $\sum_{k=0}^{n} \frac{e^{\sum_{i=0}^kx_{i}}}{\sum_{i=0}^k x_{i}}$
Is it possible to simplify
$$
S(\mathbf{x})=\sum_{k=0}^{n} \frac{e^{\sum_{i=0}^kx_{i}}}{\sum_{i=0}^k x_{i}}
$$
A few observations:
$\sum_{k=0}^{n}\sum_{i=0}^k x_{i}=\sum_{k=0}^{n}(n+1-k)x_k$
$ e^{\...
- 3,435
2
votes
1
answer
107
views
$a_{n+1}=\ln\sum_{k=1}^n a_k$
Let $(a_n)_{n\ge1}$ be a sequence of strictly positive numbers defined by the recursion $$a_{n+1}=\ln\sum_{k=1}^n a_k$$ The problem asks to calculate $$\lim_{n\to\infty} \bigg(\frac{a_{n+1}}{a_n}-1\...
- 741
5
votes
2
answers
244
views
Evaluating the limit $\lim_{n \to \infty} \left[\log(n) + 2n \log(4n + 2) - \sum_{k = 1}^{2n} (-1)^k (2k+1)\log(2k+1)\right]$
I am trying to show that the limit
$$\lim_{n \to \infty} \left[\log(n) + 2n \log(4n+2) - \sum_{k=1}^{2n} (-1)^k (2k+1) \log(2k + 1)\right] = \frac{2G}{\pi} - \log(4)$$
where $G$ is Catalan's constant.
...
2
votes
0
answers
39
views
Simplifying $\sum_{k=1}^{\log_{2}{n}}\log_{3}{\frac{n}{2^k}}$ in two ways gives different results
I want to calculate the result of
$$\sum\limits_{k=1}^{\log_{2}{n}}\log_{3}{\frac{n}{2^k}}$$ I used two below approaches. Both approaches are based on $\log A + \log B = \log (A \times B)$ and $\sum\...
- 485