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algebra precalculus - Evaluating $\frac {1} {a^ {2025}}+\frac {1} {b . . .
Well, the image equation is a different equation? One has $\frac1 {2024}$ on the right, and the other has $2024$ on the right?
calculus - Evaluating $\int {\frac {x^ {14}+x^ {11}+x^5} { (x^6+x^3+1 . . .
The following question is taken from JEE practice set Evaluate $\displaystyle\int {\frac {x^ {14}+x^ {11}+x^5} {\left (x^6+x^3+1\right)^3}} \, \mathrm dx$ My
Evaluating $\\lim_{n\\to\\infty}\\left( \\frac{\\cos\\frac{\\pi}{2n . . .
Since the OP solve his her problem, I just as well complete the solution: \begin {align} \frac {1} {n+1}\sum^n_ {k=1}\cos\left (\tfrac {k\pi} {2n}\right) =\frac {n
Evaluating $ \lim\limits_ {n\to\infty} \sum_ {k=1}^ {n^2} \frac {n} {n . . .
How would you evaluate the following series? $$\\lim_{n\\to\\infty} \\sum_{k=1}^{n^2} \\frac{n}{n^2+k^2} $$ Thanks
Evaluating $\\prod_{n=1}^{\\infty}\\left(1+\\frac{1}{2^n}\\right)$
Compute:$$\prod_ {n=1}^ {\infty}\left (1+\frac {1} {2^n}\right)$$ I and my friend came across this product Is the product till infinity equal to $1$? If no, what is the answer?
Evaluating $\sum_ {i=1}^ {\infty}\frac { (i\ln 2)^i} {2^ii!}$
I seek the proof of the evaluation to the sum $$\sum_ {i=1}^ {\infty}\frac { (i\ln 2)^i} {2^ii!} = \frac {1} {1-\ln2}-1 \approx 2 25889 $$ It is almost a power series