# conegrens of sine mmn11 2a

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• Mar 3rd 2011, 01:22 AM
transgalactic
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)
??
• Mar 3rd 2011, 03:52 AM
Ted
Try to use the limit comparison test with $\displaystyle \displaystyle b_k=\dfrac{1}{k}$
• Mar 3rd 2011, 04:39 AM
chisigma
Quote:

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

$\displaystyle \chi$ $\displaystyle \sigma$
• Mar 3rd 2011, 05:10 AM
transgalactic
i did compared it to 1/k
but 1/k is diverging

so it doesnt say anything about it
• Mar 3rd 2011, 06:01 AM
chisigma
Quote:

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 $\displaystyle \displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} = \ln 2$...
• Mar 3rd 2011, 06:12 AM
transgalactic
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)
??