Calculus Question?

Question

Determine whether the following series converge or diverge. If they converge, find their sum if possible.

1/(2*3) - 3/(3*4) + 5/(4*5) - 7/(5*6) + ........................

Answer 1

It seems that your series is [k=2, ∞]∑(-1)^k (2k-3)/(k(k+1)). Then this is an alternating series, and since [k→∞]lim (2k-3)/(k(k+1)) = 0, this series must converge by the alternating series test.

To find the sum, we must first find a closed-form expression for the partial sums. Consider the partial fraction expansion:

(2k-3)/(k(k+1)) = A/k + B/(k+1)
2k-3 = A(k+1) + Bk

Setting k=0 yields A = -3 and k=-1 yields B = 5. So we have:

[k=2, ∞]∑(-1)^k (5/(k+1) - 3/k)
5 [k=2, ∞]∑(-1)^k/(k+1) - 3[k=2, ∞]∑(-1)^k/k

Re-indexing and making other small changes:

5 [k=3, ∞]∑(-1)^(k-1)/k - 3[k=2, ∞]∑(-1)^k/k
5 [k=3, ∞]∑(-1)^(k+1)/k + 3[k=2, ∞]∑(-1)^(k+1)/k

Now, we have the well-known Mercator series for the natural logarithm -- ln (1+x) = [k=1, ∞]∑(-1)^(k+1) x^k/k. In particular, [k=1, ∞]∑(-1)^(k+1)/k = ln 2. So we have:

5 ([k=1, ∞]∑(-1)^(k+1)/k - (1 - 1/2)) + 3([k=1, ∞]∑(-1)^(k+1)/k - 1)
5 (ln 2 - (1 - 1/2)) + 3(ln 2 - 1)
5 ln 2 - 5/2 + 3 ln 2 - 3
8 ln 2 - 11/2 ≈ 0.0451774445
And we are done.

Answer 2

It converges by alternative series test.