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Exercise 5.5 · Q20

Q.The value of 1−12(23)+13(23)2−14(23)3+⋯1-\dfrac12\left(\dfrac23\right)+\dfrac13\left(\dfrac23\right)^2-\dfrac14\left(\dfrac23\right)^3+\cdots is

(1) log⁡(53)\log\left(\dfrac53\right)
(2) 32log⁡(53)\dfrac32\log\left(\dfrac53\right)
(3) 53log⁡(53)\dfrac53\log\left(\dfrac53\right)
(4) 23log⁡(23)\dfrac23\log\left(\dfrac23\right)
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Recognise the series as a rescaled logarithmic series with y=23y=\frac23, factor out the leading yy to match the standard ∑(−1)k−1yk/k\sum(-1)^{k-1}y^k/k form, and evaluate.

Step 1. Write the general term. The series is ∑k=1∞(−1)k−11k(23)k−1\displaystyle\sum_{k=1}^\infty (-1)^{k-1}\frac1k\left(\frac23\right)^{k-1} (checking: k=1k=1 gives 11; k=2k=2 gives −12⋅23-\frac12\cdot\frac23; k=3k=3 gives +13(23)2+\frac13\left(\frac23\right)^2 — matches the given series).

Step 2. Rewrite to match the standard logarithmic series form ∑(−1)k−1ykk=log⁡(1+y)\sum(-1)^{k-1}\frac{y^k}k=\log(1+y). Factor out 12/3=32\dfrac1{2/3}=\dfrac32:

∑k=1∞(−1)k−11k(23)k−1=32∑k=1∞(−1)k−11k(23)k.\sum_{k=1}^\infty(-1)^{k-1}\frac1k\left(\frac23\right)^{k-1} = \frac32\sum_{k=1}^\infty(-1)^{k-1}\frac1k\left(\frac23\right)^k. …

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