find-the-sum-n-1-1-n-2-3-n-

Question Number 89313 by abdomathmax last updated on 16/Apr/20

f i n d t h e s u m \sum_{n = 1}^{\infty} \frac{1}{n^{2} \times 3^{n}}

Commented by mathmax by abdo last updated on 19/Apr/20

l e t f i n d s (x) = \sum_{n = 1}^{\infty} \frac{x^{n}}{n^{2}} w i t h ∣ x ∣< 1

w e h a v e \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} \Rightarrow \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n + 1} = - l n (1 - x) \Rightarrow

\sum_{n = 1}^{\infty} \frac{x^{n}}{n} = - l n (1 - x) \Rightarrow \sum_{n = 1}^{\infty} \frac{x^{n - 1}}{n} = - \frac{l n (1 - x)}{x} \Rightarrow

\sum_{n = 1}^{\infty} \frac{x^{n}}{n^{2}} = - \int \frac{l n (1 - x)}{x} d x + c = - \int_{0}^{x} \frac{l n (1 - t)}{t} d t + c

s (x) = c - \int_{0}^{x} \frac{l n (1 - t)}{t} d t

s (0) = 0 = c \Rightarrow s (x) = - \int_{0}^{x} \frac{l n (1 - t)}{t} d t \Rightarrow

\sum_{n = 1}^{\infty} \frac{1}{n^{2} \times 3^{n}} = s (\frac{1}{3}) = - \int_{0}^{\frac{1}{3}} \frac{l n (1 - t)}{t} d t

=_{u = 3 t} - 3 \int_{0}^{1} \frac{l n (1 - \frac{u}{3})}{u} \times \frac{d u}{3} = - \int_{0}^{1} \frac{l n (1 - \frac{u}{3})}{u} d u

\frac{u}{3} = c o s θ g i v e Σ (\dots) = - \int_{\frac{π}{2}}^{a r c o s (\frac{1}{3})} \frac{l n (1 - c o s θ)}{3 c o s θ} (- 3 s i n θ) d θ

= \int_{\frac{π}{2}}^{a r c o s (\frac{1}{3})} t a n θ l n (2 {s i n}^{2} (\frac{θ}{2})) d θ

= l n (2) \int_{\frac{π}{2}}^{a r c o s (\frac{1}{3})} t a n θ d θ + 2 \int_{\frac{π}{2}}^{a r c o s (\frac{1}{3})} t a n θ l n (s i n (\frac{θ}{2})) d θ

{\int_{\frac{π}{2}}^{a r c o s (\frac{1}{3})} t a n θ d θ = - l n ∣ c o s θ ∣]}_{\frac{π}{2}}^{a r c o s (\frac{1}{3})} = - l n (\frac{1}{3}) = l n (3)

\dots . b e c o n t i n u e d \dots .