All Questions
24
questions
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
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
3
votes
3
answers
189
views
To compute $\lim\limits_{n\to +\infty} \log_2\left(\frac{n^3}{n^4+2}\right)=-\infty, \lim\limits_{x\to+\infty}\log_2\left(\frac{x^3}{x^4+2}\right)$
On my textbook of Maths for students of my high school (there are really only two such limits) I have found, for example, a limit of a succession with this solution:
$$\color{magenta}{\lim_{n\to +\...
- 7,792
3
votes
3
answers
86
views
Find $\log_e3 - \frac{\log_e9}{2^2} + \frac{\log_e27}{3^2} - \frac{\log_e81}{4^2} + ...$
Find $\log_e3 - \dfrac{\log_e9}{2^2} + \dfrac{\log_e27}{3^2} - \dfrac{\log_e81}{4^2} + ...$
What I Tried:- This is the same as :-
$$\ln3 - \frac{\ln3}{2} + \frac{\ln3}{3} - \frac{\ln3}{4} + \dots$$
$$...
- 4,254
7
votes
1
answer
139
views
Trying to find a NICE form of : $\sum_{m=1}^{n}\lfloor\log_2m\rfloor$ [ Mathematical Gazette 2002 ]
$Q.$ Find a NICE form of : $$\sum_{m=1}^{p}\lfloor\log_2m\rfloor$$
APPROACH : We have , $$\lfloor\log_21\rfloor⠀\color{red}{\lfloor\log_22\rfloor}⠀\lfloor\log_23\rfloor⠀\color{red}{\lfloor\log_24\...
- 688
4
votes
4
answers
633
views
(AIME 1994) $ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $
$($AIME $1994)$ Find the positive integer $n$ for which
$$ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \lfloor \log_2 3 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $$ where $\lfloor ...
- 41
1
vote
3
answers
66
views
Sum of a series given the formula for the nth term
So I have a series $1+1+3+9+23+\cdots$ and the formula given to find the $n^{th}$ term $R_n=1-2n+2^n$. The first part is to verify that the term $115$ exists in this series. I knew that if I just ...
- 47
1
vote
2
answers
86
views
Explanation of this Derivation of the Logarithmic Series
I have question on a derivation of the logarithmic series that I'm still unable to understand. I've already seen this post on it, but I'm still slightly confused. This derivation is from Higher ...
- 543
2
votes
3
answers
110
views
Find the sum $\sum_{n=1}^{\infty} (-1)^{n-1}\log \left(1- \frac{1}{(n+1)^2}\right)$ based on $a^2-b^2 $?
Find the sum of the given series
$$\sum_{n=1}^{\infty} (-1)^{n-1}\log \left(1- \frac{1}{(n+1)^2}\right).$$
My attempt : I used the formula $a^2- b^2 = (a+b)(a-b)$, then I get
$$\sum_{n=1}^{\infty}...
- 14.6k
0
votes
0
answers
64
views
Intersection between an AP and GP (solving equation)
Given the following equation:
$$1000+(N-1)200=500\cdot 1.2^N-1$$
I need to find the value for $N$.
When there was no $N$ in the left-hand-side I will be able to solve the equation with logarithms. ...
- 11
0
votes
1
answer
368
views
An arithmetic progression of logarithms [duplicate]
$1, \log_yx, \log_zy, -15\log_xz$ are in arithmetic progression, then which of the following are correct:
$z^3 = x$
$x = y^{-1}$
$z^{-3} = y$
$x = y^{-1} = z^3$
I tried converting the logs into a ...
- 359
2
votes
2
answers
68
views
Is it ok to switch to logarithms when testing divergence of sequences involving a sum of exponentials? E.g $x_n = 5^n - 4^n \to \ln5^n - \ln4^n$
Consider the following sequence:
$$
x_n = 5^n - 4^n
$$
I'm wondering whether it is ok to switch from comparing exponents to comparing logarithms. Here is what I mean:
Instead of $5^n - 4^n$ compare:...
- 5,411
4
votes
5
answers
201
views
Verify the proof that $x_n = \ln^2(n+1) - \ln^2n$ is a bounded sequence.
Let $n\ \in \mathbb N$ and:
$$
x_n = \ln^2(n+1) - \ln^2n
$$
Prove that $x_n$ is a bounded sequence.
I've taken the following steps. Consider $x_n$
$$
\begin{align}
x_n &= \ln^2(n+1) - \ln^2n =...
- 5,411
4
votes
5
answers
150
views
Solve: $2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1\Bigl) = 0$
The question says to find the value of $x$ if, $$2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl)= 0$$
My approach:
I rewrote the expression as,
$$2^x\...
- 1,312
2
votes
2
answers
157
views
Computing $\lim\limits_{n\to \infty} \frac{\ln^2n}{n}\sum\limits_{k=2}^{n-2}\frac{1}{\ln k\ln(n-k)}$
Computing the following limit $$\lim_{n\to \infty} \frac{\ln^2n}{n}\sum_{k=2}^{n-2}\frac{1}{\ln k\ln(n-k)}$$
I first tried to write it as a Riemann sum as follows
$$\frac{\ln^2n}{n}\sum_{k=2}^{n-2}\...
- 24.2k