More series issues!

for #10, i have found that when p < = 0, the series converges. fact check on that?

It can't be that makes the series converge, because this would introduce a zero in the denominator when . For a similar reason, would involve an indeterminate term of .

That leaves checking what happens when . First, consider the function



and its derivative



has critical points at and . (These never coincide because we're assuming , so it's always the case that .) Between these two points, say at , you have , which is positive regardless of the value of . This means is increasing on the interval .

Meanwhile, if - and let's take as an example - we have , which is negative for all . This means is decreasing for all , which shows that is a decreasing sequence for all , where is any sufficiently large number that depends on .

Now, it's also the case that for (and in fact all ),



So you have a series of a sequence that in absolute value is decreasing and converging to 0. The alternating series then says that the series must converge for all .

For the second question, recall that




(note that the above is true for even only - it wouldn't be too difficult to change things around if is odd)

It follows that




Subtracting from , you have







as required. Notice that assuming is odd doesn't change the result; the last term in ends up canceling with the corresponding term in regardless of the parity of .

5/6, -1/6, 5/3 i hope this is right. it should be anyway.