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Homework Statement
Qn 1:
[tex]\sum_{k=1}^{\infty}(-1)^{k}((\sqrt{k}+1)/(k+1))[/tex]
Interval of convergence for the series:
[tex]\sum_{k=0}^{\infty}((x^{k})/(\sqrt{(k²+3)}))[/tex]
The ans is [-1,1]
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Thanks Gib for the great hints.For the first one - it is an alternating series. Have you tried to apply the Leibniz criterion?
For the second one - use the ratio test. And the answer isn't [-1,1].
Uhh... Sorry Gib.. please help me on the first one agn.. I've made typo earlier.If I am reading correctly, is the term in your first one [tex] \frac{ \sqrt{k+1}}{k+1} = \frac{1}{\sqrt{k+1}} [/tex] ? If so, it should be evident that it is decreasing, but you can also check the difference of consecutive terms is negative quite easily.
For the second one, you have the right answer. The inside interval (-1,1) follows from the ratio test, the endpoint -1 follows from the Leibniz criterion, and the endpoint 1 is dealt with by comparison to a well known divergent series. Your lecturer probably made a silly mistake, and will be impressed if you also add in the proof of why it is divergent at x=1.
Oh! thanks! I was reluctant i split into seperate sums ...You can see that [tex] \sum (-1)^k \frac{1}{k+1} [/tex] converges, so we need only check [tex] \sum (-1)^k \frac{ \sqrt{k}}{k+1} [/tex] converges or diverges, since by addition we get the original series back (of course, we use "Convergent + Convergent = Convergent, Convergent + Divergent = Divergent").
So for the simpler one we are checking, by the Leibniz criterion we must check if the term is eventually decreasing. One way to do this is to show the ratio of consecutive terms is less than 1, so try to show that.
But we can only show the ratio is less than 1 for finite terms is it sufficient?Don't do comparison, just apply Leibniz straight away. 1/(k+1) is decreasing.
Show the ratio of consecutive terms is less than 1 for [tex] \sum (-1)^k \frac{\sqrt{k}}{k+1} [/tex] too.