All Questions
31
questions
3
votes
4
answers
82
views
Finding and proofing a closed formula for $\sum_{n=1}^k\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}$
I want to find and proof a closed formula for the following sum $$\sum_{n=1}^k\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\dots +\...
- 524
-1
votes
1
answer
126
views
Find the value $\sum_{n=a}^b\frac1{\sin (2^{n+3})}$ [closed]
Find the value of: $$\sum_{n=0}^{10}\frac1{\sin (2^{n+3})}$$
I'm stuck on this problem, can someone please help?
- 596
-1
votes
1
answer
75
views
how to find the sum of these terms without the gamma function?
While solving a problem based on integration, I arrived at the following
$$\sum\limits_{x = 1}^{38} \ln\left(\frac{x}{x+1}\right)$$
I'm supposed to prove that this is less than $\ln(99)$
in order to ...
- 1,326
2
votes
1
answer
88
views
Best way to solve a summation with binomial coefficients in denominator apart from telecoping method
The value of $\sum_{r=1}^{m}\frac{(m+1)(r-1)m^{r-1}}{r\binom{m}{r}} = \lambda$ then the correct statement is/are
(1) If $m=15$ and $\lambda$ is divided by m then the remainder is 14.
(2) If $m=7$ and $...
- 445
-2
votes
2
answers
242
views
Sum the series : $\frac{1}{9\sqrt11 + 11\sqrt9} +\frac{1}{11\sqrt13 + 13\sqrt11} +\ldots$ [closed]
$$\frac{1}{9\sqrt11 + 11\sqrt9} + \frac{1}{11\sqrt13 + 13\sqrt11} + \frac{1}{13\sqrt15 + 15\sqrt13} + \ldots + \frac{1}{n\sqrt{n+2} + (n+2)\sqrt{n}} = \frac{1}{9}$$
Find the value of $n$.
I got the ...
- 187
3
votes
1
answer
144
views
Find $\sum_{r=1}^{20} (-1)^r\frac{r^2+r+1}{r!}$.
Calculate $$\sum_{r=1}^{20} (-1)^r\frac{r^2+r+1}{r!}\,.$$
I broke the sum into partial fractions and after writing 3-4 terms of the sequence I could see that it cancels but I wasn't able to arrive at ...
- 1,107
1
vote
1
answer
143
views
Evaluate $\sum_{r=1}^{m} \frac{(r-1)m^{r-1}}{r\cdot\binom{m}{r}}$
Evaluate:$$\sum_{r=1}^{m} \frac{(r-1)m^{r-1}}{r\cdot\binom{m}{r}}$$
Using the property:$$r\binom{m}{r}=m\binom{m-1}{r-1}$$
It is same as $$\sum_{r=2}^{m} \frac{(r-1)m^{r-1}}{m\cdot\binom{m-1}{r-1}}$$
...
- 868
0
votes
3
answers
2k
views
If $\sum_{r=0}^{n-1}\log _2\left(\frac{r+2}{r+1}\right)= \prod_{r = 10}^{99}\log _r(r+1)$, then find $n$.
If \begin{align}\sum_{r=0}^{n-1}\log _2\left(\frac{r+2}{r+1}\right) = \prod_{r = 10}^{99}\log _r(r+1).\end{align}
then find $n$.
I found this question in my 12th grade textbook and I just can't wrap ...
- 125
1
vote
1
answer
183
views
Find the sum: $\sum_{n=1}^{20}\frac{(n^2-1/2)}{(n^4+1/4)}$
Hint: this is a telescoping series sum (I have no prior knowledge of partial fraction decomposition)
Attempt: I tried to complete the square but the numerator had an unsimplifiable term. So I couldn't ...
- 45
2
votes
3
answers
623
views
Prove sum of $k^2$ using $k^3$
So the title may be a little bit vague, but I am quite stuck with the following problem.
Asked is to first prove that $(k + 1)^3 - k^3 = 3k^2 + 3k + 1$. This is not the problem however. The question ...
- 181
1
vote
2
answers
83
views
Finding a formula for $\sum_{k=1}^n(k^2-(k-1)^2)$
I have got this following series:
$$\sum_{k=1}^n(k^2-(k-1)^2)$$
I want to come up with a formula for the summation.
I did some math and for me, the formula would be as follows:
$$\sum_{k=1}^n(k^2-(...
- 159
0
votes
2
answers
184
views
Possible telescopic sum
Prove that $$\sum_{k=1}^n 4^{k}\sin^{4} \left(\frac{a}{2^k}\right) = 4^{n}\sin^{2} \left(\frac{a}{2^n}\right) - \sin^{2}a$$
I suspect that telescopic sum is involved but don't know how to proceed. ...
- 407
4
votes
2
answers
387
views
Find the sum of series: $\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{97}+\sqrt{98}}+\frac{1}{\sqrt{99}+\sqrt{100}}$
Find the sum of series:
$$\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{97}+\sqrt{98}}+\frac{1}{\sqrt{99}+\sqrt{100}}$$
My Attempt:
I tried ...
- 9,569
0
votes
2
answers
52
views
Rewrite $\sum_{n=1}^k\log_3{(\frac{n+1}{n})}$ and write the formula in terms of k
Rewrite $\sum_{n=1}^k\log_3{(\frac{n+1}{n})}$ and write the formula in terms of k.
I rewrote to $1+\frac{1}{n}$ and summed to get (I think) $\log_3(k+\frac{1}{n^k+k!})$ but I'm unsure if the $\log_3$ ...
- 37
0
votes
2
answers
56
views
Rewrite $\sum_{n=1}^k{(n-1)/n!}$ and write the formula in terms of k [closed]
Rewrite $\sum_{n=1}^k{\frac{n-1}{n!}}$
I have turned it into $\frac{1}{n}*\frac{1}{(n-2)!}$ but do not know where to go from here.
- 37