# Thread: having a problem in determine the Convergence or divergence for infinite sum.

1. ## having a problem in determine the Convergence or divergence for infinite sum.

can anyone show me how to
Determine whether ∑(k=1 to n) [(1-ksin(1/k)]
converges .
TQ

2. If the general term of the series is $\displaystyle a_{k} = 1- \sin \frac{1}{k}$ the series diverges because is $\displaystyle \lim_{k \rightarrow \infty} a_{k} = 1 \ne 0$... may be that the $\displaystyle a_{k}$ are something else?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. Originally Posted by chisigma
If the general term of the series is $\displaystyle a_{k} = 1- \sin \frac{1}{k}$ the series diverges because is $\displaystyle \lim_{k \rightarrow \infty} a_{k} = 1 \ne 0$... may be that the $\displaystyle a_{k}$ are something else?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

oh yea.. mistype a k there.. updated.. tQ.

4. All right!... so the series is...

$\displaystyle \sum_{k=1}^{\infty} (1-k\cdot \sin \frac{1}{k})$ (1)

Using the series expansion...

$\displaystyle \sin x= \sum_{n=0}^{\infty} \frac{(-1)^{n}\cdot x^{2n+1}}{(2n+1)!}$ (2)

... we find that...

$\displaystyle 1 - k\cdot \sin \frac{1}{k} = - \sum_{n=1}^{\infty}\frac{(-1)^{n}}{k^{2n}\cdot (2n+1)!}$ (3)

... so that the (1) becomes...

$\displaystyle \sum_{k=1}^{\infty} (1-k\cdot \sin \frac{1}{k})= \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{k^{2n}\cdot (2n+1)!}=$

$\displaystyle = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}\cdot \zeta (2n)}{(2n+1)!} < 2\cdot \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n+1)!} = 2\cdot (1-\sin 1)$ (4)

... and the series (1) converges...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. Originally Posted by chisigma
All right!... so the series is...

$\displaystyle \sum_{k=1}^{\infty} (1-k\cdot \sin \frac{1}{k})$ (1)

Using the series expansion...

$\displaystyle \sin x= \sum_{n=0}^{\infty} \frac{(-1)^{n}\cdot x^{2n+1}}{(2n+1)!}$ (2)

... we find that...

$\displaystyle 1 - k\cdot \sin \frac{1}{k} = - \sum_{n=1}^{\infty}\frac{(-1)^{n}}{k^{2n}\cdot (2n+1)!}$ (3)

... so that the (1) becomes...

$\displaystyle \sum_{k=1}^{\infty} (1-k\cdot \sin \frac{1}{k})= \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{k^{2n}\cdot (2n+1)!}=$

$\displaystyle = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}\cdot \zeta (2n)}{(2n+1)!} < 2\cdot \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n+1)!} = 2\cdot (1-\sin 1)$ (4)

... and the series (1) converges...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

TQ.Just wanna share with u, i found another alternative way of solving this question and it is much more simpler .By using limit comparison test comparing to 1/k^2 .