How to tell what my convergent Infinite Series actually converges to?

Question

I have an infinite series which is convergent and I want to know exactly what number it converges to, to very high precision. Is this possible and if so, by what methods? The series in question is:

Sum for n =0 to infinite of 1/[(n+1)(1+n/2)(2^(2n+2))]

Answer 1

The answer is 6 ln (3/4) + 2 = 0.273908...

To see this:
1/[(n+1)(1+n/2)(2^(2n+2))] =
1/[(n+1)(n+2)(2^(2n+1))] =
1/[(n+1)(2^(2n+1))] - 1/[(n+2)(2^(2n+1))]
where the last step is partial fractions.

Now we'll do each sum individually.

sum for n=0 to infinity of 1/[(n+1)(2^(2n+1))] =
sum for n=0 to infinity of 2/[(n+1)(2^(2n+2))] =
sum for n=0 to infinity of 2/[(n+1)(4^(n+1))] =
sum for m=1 to infinity of 2/(m4^m)

Use the series -ln(1-x) = x+x^2/2+x^3/3+x^4/4... to get that the above sum is
-2 ln (3/4).

Similarly,

sum for n=0 to infinity of 1/[(n+2)(2^(2n+1))] =
sum for n=0 to infinity of 8/[(n+2)(2^(2n+4))] =
sum for n=0 to infinity of 8/[(n+2)(4^(n+2))] =
sum for k=2 to infinity of 8/(k4^k) =
8 * -(ln (3/4) - 1/4). The 1/4 part came from the fact that the sum starts at 2.

So the overall sum is -2 ln (3/4) - 8 * (-ln (3/4) - 1/4) = 6 ln (3/4) + 2.

Hope this helps.

Answer 2

Neither (-a million)^n nor n*(-a million)^n are convergent infinite series. an major (yet no longer adequate) situation for any series to converge is that the series of words ought to bypass to 0, and neither of those fulfill that requirement. i ask your self in case you meant to ask even if (a million/n)*(-a million)^n converges. actually it does. there's a regular theorem which states that an alternating series (one the position the signal of the words alternates between effective and unfavourable) converges every time the series of absolutely the values of the words decreases monotonically to 0. For an celebration of a series that has radius of convergence a million, yet diverges at -a million and a million use the geometric series (out of your question i wager this can be what you already did). Radius of convergence a million and convergence at a million yet no longer -a million is extra sturdy, even if i imagine that (the McLauren series for) ln(a million+x) works.

Answer 3

42