n-1-m-1-m-2-n-3-m-3-n-m-3-m-n-find-

Question Number 192590 by York12 last updated on 22/May/23

\underset{n = 1}{\sum^{\infty}} \underset{m = 1}{\sum^{\infty}} [\frac{m^{2} n}{3^{m} (3^{n} . m + 3^{m} . n)}] = λ

f i n d λ .

Answered by witcher3 last updated on 24/May/23

(n, m) \overset{f}{\to} \frac{m^{2} n}{3^{m} (3^{n} m + 3^{m} n)}

\forall n \in Z_{+}, \sum_{m ⩾ 1} f (n, m) = n \sum_{m ⩾ 1} \frac{m^{2}}{3^{m} (3^{n} m + 3^{m} n)}

⩽ n \sum_{m ⩾ 1} \frac{m^{2}}{3^{m}} = S

\exists N \in N \forall m ⩾ N, m^{2} ⩽ 2^{m}

S ⩽ \underset{m = 1}{\sum^{N - 1}} n \frac{m^{2}}{3^{m}} + n \sum_{N} {(\frac{2}{3})}^{m} < \infty

\forall m \in Z_{+} f (n, m) = \frac{m^{2}}{3^{m}} Σ \frac{n}{(3^{n} m + 3^{m} n)} ⩽ \frac{m^{2}}{3^{m}} Σ \frac{n}{3^{n}}

Sam as previous < \infty

\Rightarrow \sum_{n} \sum_{m} = \sum_{m} \sum_{n}

Σ Σ f (n, m) = Σ Σ f (m, n) = S

2 S = \sum_{n} \sum_{m} \frac{m^{2} n}{3^{m} ({n 3}^{m} + m 3^{n})} + \frac{n^{2} m}{3^{n} (n 3^{m} + m 3^{n})}

= \sum_{n} \sum_{m} \frac{mn ({m 3}^{n} + {n 3}^{m})}{3^{m + n} ({n 3}^{m} + {m 3}^{n})} = \sum_{n} \frac{n}{3^{n}} Σ \frac{m}{3^{m}}

\frac{1}{{(1 - x)}^{2}} = \sum_{n ⩾ 1} {nx}^{n - 1} \Rightarrow \frac{x}{{(1 - x)}^{2}} = Σ {nx}^{n}

S = {(\frac{\frac{1}{3}}{{(1 - \frac{1}{3})}^{2}})}^{2} = \frac{9}{16}

Commented by York12 last updated on 24/May/23

G r e a t!

w h a t b o o k d o y o u r e c o m m e n d f o r a l g e b r a

Commented by witcher3 last updated on 24/May/23

hello sir sorry for last times

not responding i have not telegram

book 1 introduce to matrix and linear algebra

2) for calculus

almost Impossible sum and integrals

contein hards problems!

Commented by York12 last updated on 24/May/23

t h a n k s s o m u c h, N o p r o b l e m

Facebook Tweet Pin