# Thread: conegrens of sine mmn11 2a

1. ## conegrens of sine mmn11 2a

is it converges absolutly ,or conditionally convergent
or is it diverges?

$\displaystyle \sum_{k=1}^{infinity}(-1)^{k}sin\frac{1}{k}$
the absolute value series is |sin(1/k)|<=|1/k|
which is diverges

but it gives me nothing
because if it where convergent then the smaller one would converge
but here the bigger one diverges(harmonic series)
??

2. Try to use the limit comparison test with $\displaystyle \displaystyle b_k=\dfrac{1}{k}$

3. Originally Posted by transgalactic
is it converges absolutly ,or conditionally convergent
or is it diverges?

$\displaystyle \sum_{k=1}^{infinity}(-1)^{k}sin\frac{1}{k}$
the absolute value series is |sin(1/k)|<=|1/k|
which is diverges

but it gives me nothing
because if it where convergent then the smaller one would converge
but here the bigger one diverges(harmonic series)
??
Pauls Online Notes : Calculus II - Alternating Series Test

Kind regards

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

4. i did compared it to 1/k
but 1/k is diverging

so it doesnt say anything about it

5. Originally Posted by transgalactic
i did compared it to 1/k
but 1/k is diverging

so it doesnt say anything about it
Of course $\displaystyle \displaystyle \sum_{k=1}^{\infty} \frac{1}{k}$ diverges... but also is $\displaystyle \displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} = \ln 2$...

6. first we need to compare the absolute value
the absolute value series is |sin(1/k)|<=|1/k|
which is diverges

but it gives me nothing
because if it where convergent then the smaller one would converge
but here the bigger one diverges(harmonic series)
??