If S-1-3-1-32-1-33- up to - then-

S= 1/3+1/3^2+1/3_3 —It is an Gp with a= 1/3 amp;r 1/3 so = a/1 r=1/3/1 1/3=1/3/2/3=1/2 ...

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Summation by Sigma Method

if S=1/3+1/...

Question

if $S = \frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + . . .$ up to $\infty$ , then:

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1 \cdot 3 + 3 \cdot 3^{2} + 5 \cdot 3^{3} + 7 \cdot 3^{4} + \dots

n

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t

S_{\infty} = 1 + \frac{4}{3} + \frac{9}{3^{2}} + \frac{16}{3^{3}} + \dots \dots

2 S_{\infty}

{(1 + \frac{2}{3} + \frac{6}{3^{2}} + \frac{10}{3^{3}} + \dots upto \infty)}^{{log}_{(0.25)} ⎛ ⎝ \frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \dots upto \infty ⎞ ⎠}

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(0. 64)^{l o g_{0.25} ⎛ ⎝ \frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} . . . . . \infty ⎞ ⎠}

S = \frac{1}{3} + \frac{2}{3^{2}} + \frac{3}{3^{3}} + \frac{4}{3^{4}} + \dots \infty,

4 S

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