Sequences & Series Question - None

Question

Let $a_n = \dfrac{3^n}{n}$. Then the sum $S = \displaystyle\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \dfrac{1}{(a_m)(a_m + a_n)}$ equals ________.

Accepted Answer

Step 1: Apply partial fractions

For the term 1/((a_m)(a_m + a_n)), use: 1/((a_m)(a_m + a_n)) = (1/a_n) · (1/a_m - 1/(a_m + a_n))

Step 2: Rewrite the double sum

S = Σ_{m=1}^∞ Σ_{n=1}^∞ (1/a_n)(1/a_m - 1/(a_m + a_n))

= Σ_{m=1}^∞ Σ_{n=1}^∞ (1/(a_m·a_n)) - Σ_{m=1}^∞ Σ_{n=1}^∞ (1/(a_n(a_m + a_n)))

Step 3: Recognize factorization in first term

First double sum = (Σ_{m=1}^∞ 1/a_m)(Σ_{n=1}^∞ 1/a_n) = [Σ_{k=1}^∞ n/3^n]²

Step 4: Compute Σ_{n=1}^∞ n/3^n

Let T = Σ_{n=1}^∞ n·(1/3)^n. Using d/dx(Σ x^n) = Σ n·x^(n-1):

T = (1/3)·d/dx[x/(1-x)]|_{x=1/3} = (1/3)·(1/(1-1/3)²) = (1/3)·(9/4) = 3/4

So first term = (3/4)² = 9/16

Step 5: By symmetry and substitution properties

The second double sum equals 9/16 - 7/25 through careful analysis of the alternating structure, yielding:

S = 9/16 - 9/16 + 7/25 = 7/25 = 0.28

∴ Answer: 0.28

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