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Sum of Series 4/19 + 44/19^2 + 444/19^3 + ... | JEE

JEE Maths question with a full step-by-step solution.

Question
The sum of the series 419+44192+444193+⋯\frac{4}{19}+\frac{44}{19^2}+\frac{444}{19^3}+\cdots to infinity is
A3881\frac{38}{81}correct
B419\frac{4}{19}
C36371\frac{36}{371}
D1819\frac{18}{19}
Solution
Step 1: Let SS be the sum and form S−S19S-\dfrac{S}{19}:
S−S19=419+40192+400193+⋯ .S-\frac{S}{19}=\frac{4}{19}+\frac{40}{19^2}+\frac{400}{19^3}+\cdots.
 Step 2: The right side is a G.P. with first term 419\dfrac{4}{19} and ratio 1019\dfrac{10}{19}:
18S19=419⋅11−1019=419⋅199=49.\frac{18S}{19}=\frac{4}{19}\cdot\frac{1}{1-\frac{10}{19}}=\frac{4}{19}\cdot\frac{19}{9}=\frac49.
 Step 3: So S=49⋅1918=3881.S=\dfrac49\cdot\dfrac{19}{18}=\dfrac{38}{81}. Correct answer: (1)
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