More proofs involving series?

you already seem to have the answer to 1), so I'll be skipping that.

for 2), you are definitely on the right track, though it is not necessary to show that the remainder is 0 once you've integrated both sides.

we have:

∑(n=0 to inf) (-1)^n*t^2n = 1/(1+t^2)

integrating both sides (from 0 to x with respect to t), we have

∫(0 to x) (-1)^n*t^2n dt = ∫(0 to x) 1/(1+t^2) dt -->

∑(n=1 to inf) (-1)^n*(x^2n+1)/(2n+1) = arctan(x) - arctan(0) = arctan(x)

thus, we have proven the equality

3) we now know that

∑(n=1 to inf) (-1)^n*(x^2n+1)/(2n+1) = arctan(x)

our goal in this problem is to change the sum we're presented into the form of some arctan(x) function.

So, looking at the solution, π/(2√3), let's try and get it in the form of something that we can take the tan of.

If we divide both sides by 1/√3, then we see that this statement is equivalent to saying

1/√3 * ∑(n=1 to inf) (-1)^n/(3^n*(2n+1)) = π/(2*3) = π/6

we know what the tangent of π/6 is, so this statement becomes:

1/√3 * ∑(n=1 to inf) (-1)^n/(3^n*(2n+1)) = arctan(1/√3)

now, change the right side into the form of a sum, and show that the remainder of the two (i.e. the limit of their difference) is zero.

A theorem may also be referred to as a lemma, especially if it is intended for use as a stepping stone in the proof of another theorem.he statement that is proved is often called a theorem.[1] Once a theorem is proved, it can be used as the basis to prove further statements.