Check if series converge?
- Thread starter ZeroXz
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\(\displaystyle \sum_{i=1}^{\infty}\frac{1}{i^{2}}\)
The above is the very famous Basel problem Euler showed converges to \(\displaystyle \frac{{\pi}^{2}}{6}\)
BTW, you have written \(\displaystyle \sum_{i=3}^{\infty}\left[sin(3i+3)+\frac{2cos(9i)}{i^{2}}\right]\)
Please do not tell me you did not use proper grouping symbols and actually meant:
\(\displaystyle \sum_{i=3}^{\infty}\frac{sin(3i+3)+2cos(9i)}{i^{2}}\)
Sorry to sound presumptuous, but this happens quite a bit from students (in calculus) who should know better.
Just checking.
The above is the very famous Basel problem Euler showed converges to \(\displaystyle \frac{{\pi}^{2}}{6}\)
BTW, you have written \(\displaystyle \sum_{i=3}^{\infty}\left[sin(3i+3)+\frac{2cos(9i)}{i^{2}}\right]\)
Please do not tell me you did not use proper grouping symbols and actually meant:
\(\displaystyle \sum_{i=3}^{\infty}\frac{sin(3i+3)+2cos(9i)}{i^{2}}\)
Sorry to sound presumptuous, but this happens quite a bit from students (in calculus) who should know better.
Just checking.
Oh sorry ! I missed out the grouping brackets which cause the ambiguity. 
It's the second version, [sin (3i + 3) + 2 cos (9i)] / i^2
So first I use comparaison test which takes me to the | [sin (3i + 3) + 2 cos (9i)] | / i^2 <= 3 / i^2
So if sigma 3/ i^2 converge, the smaller series will converge also. But which test can i use to check the 3/ i^2 converges ?
Can i use sigma 1 / n^p converges if and only if p > 1 in this case ? But the numerator is 3 instead of 1 ??
Thanks a lot
It's the second version, [sin (3i + 3) + 2 cos (9i)] / i^2
So first I use comparaison test which takes me to the | [sin (3i + 3) + 2 cos (9i)] | / i^2 <= 3 / i^2
So if sigma 3/ i^2 converge, the smaller series will converge also. But which test can i use to check the 3/ i^2 converges ?
Can i use sigma 1 / n^p converges if and only if p > 1 in this case ? But the numerator is 3 instead of 1 ??
Thanks a lot
As I said, \(\displaystyle 3\sum_{i=3}^{\infty}\frac{1}{i^{2}}\) converges because it is a very famous and known infinite series.
But, just to check convergence, the integral test would be OK.
\(\displaystyle \int_{1}^{\infty}\frac{1}{x^{2}}dx=\lim_{L\to \infty}\int_{1}^{L}\frac{1}{x^{2}}dx=\lim_{L\to \infty}\left[\frac{-1}{x}\right]_{1}^{L}=\lim_{L\to \infty}\left[1-\frac{1}{L}\right]=1\)
Since the integral converges, the series converges.
As I said, if you're interested, look up Euler's Basel problem. It shows how the sum is actually obtained.
For your case starting at k=3. It would converge to \(\displaystyle 3\left(\frac{{\pi}^{2}}{6}-\frac{5}{4}\right)\)
But, just to check convergence, the integral test would be OK.
\(\displaystyle \int_{1}^{\infty}\frac{1}{x^{2}}dx=\lim_{L\to \infty}\int_{1}^{L}\frac{1}{x^{2}}dx=\lim_{L\to \infty}\left[\frac{-1}{x}\right]_{1}^{L}=\lim_{L\to \infty}\left[1-\frac{1}{L}\right]=1\)
Since the integral converges, the series converges.
As I said, if you're interested, look up Euler's Basel problem. It shows how the sum is actually obtained.
For your case starting at k=3. It would converge to \(\displaystyle 3\left(\frac{{\pi}^{2}}{6}-\frac{5}{4}\right)\)