If y = x + x 2 2 + x 3 3 + x 4 4 + … , then show that x = y - y 2 2 ! + y 3 3 ! - y 4 4 ! + …

Given that, y = x + x 2 2 + x 3 3 + x 4 4 + ⋯ We know that, - log e ( 1 - x ) = x + x 2 2 + x 3 3 + x 4 4 + ⋯ ⇒ - y = log e ( 1 - x ) ⇒ x = 1 - e - y We know that, e y = 1 + y 1 ! + y 2 2 ! + ⋯ · ∞ = 1 - 1 - y 1 ! + y 2 2 ! - y 3 3 ! + y 4 4 ! + · · x = y - y 2 2 ! + y 3 3 ! - y 4 4 ! + · · · · · ∞ Hence, proved.

If y = x + x 2 2 + x 3 3 + x 4 4 + … , then show that x = y - y 2 2 ! + y 3 3 ! - y 4 4 ! + …

Given that, y = x + x 2 2 + x 3 3 + x 4 4 + ⋯ We know that, - log e ( 1 - x ) = x + x 2 2 + x 3 3 + x 4 4 + ⋯ ⇒ - y = log e ( 1 - x ) ⇒ x = 1 - e - y We know that, e y = 1 + y 1 ! + y 2 2 ! + ⋯ · ∞ = 1 - 1 - y 1 ! + y 2 2 ! - y 3 3 ! + y 4 4 ! + · · x = y - y 2 2 ! + y 3 3 ! - y 4 4 ! + · · · · · ∞ Hence, proved.

Show that 1 + 1 3 1 2 2 + 1 5 1 2 4 + 1 7 1 2 6 + … = log e 3 .

Given that, the series is 1 + 1 3 . 2 2 + 1 5 . 2 4 + 1 7 . 2 6 + … General term of the given series is T r = ∑ r = 0 n 1 ( 2 r + 1 ) 2 2 r We know that, log e ( 1 + x ) = x - x 2 2 + x 3 3 - x 4 4 + ⋯ and - log e ( 1 - x ) = x + x 2 2 + x 3 3 + x 4 4 + ⋯ ⇒ log e ( 1 + x ) - log e ( 1 - x ) = x - x 2 2 + x 3 3 - x 4 4 + ⋯ + x + x 2 2 + x 3 3 + x 4 4 + ⋯ Now, substitute x = 1 2 in the above expansion we get log e 1 + 1 2 - log e 1 - 1 2 = 2 1 2 + 1 2 3 3 + 1 2 5 5 + ⋯ ⇒ log e 3 2 - log e 1 2 = 1 + 1 2 2 3 + 1 2 4 5 + ⋯ ⇒ 1 + 1 3 . 2 2 + 1 5 . 2 4 + 1 7 . 2 6 + … = log e 3 Hence, proved.

Find the sum of series x - 1 x + 1 + 1 2 x 2 - 1 ( x + 1 ) 2 + 1 3 x 3 - 1 ( x + 1 ) 3 + …

Given that, x - 1 x + 1 + 1 2 x 2 - 1 ( x + 1 ) 2 + 1 3 x 3 - 1 ( x + 1 ) 3 + ⋯ . Let, S = ( x - 1 ) ( x + 1 ) + 1 2 x 2 - 1 ( x + 1 ) 2 + 1 3 x 3 - 1 ( x + 1 ) 3 + ⋯ ⇒ S = x x + 1 - 1 ( x + 1 ) + 1 2 x 2 ( x + 1 ) 2 - 1 2 ( x + 1 ) 2 + 1 3 x 3 ( x + 1 ) 3 - 1 3 × 1 ( x + 1 ) 3 + · · ⇒ S = x x + 1 + 1 2 x x + 1 2 + 1 3 x x + 1 3 + · · · - 1 ( x + 1 ) + 1 2 ( x + 1 ) 2 + 1 3 ( x + 1 ) 3 · · · We know that, log ( 1 - x ) = - x - x 2 2 - x 3 3 - x 4 4 + ⋯ ⇒ - log 1 - x x + 1 - - log 1 - 1 x + 1 = - log 1 x + 1 + log x x + 1 = log ( x + 1 ) + log x - log ( x + 1 ) = log x ⇒ x - 1 x + 1 + 1 2 x 2 - 1 ( x + 1 ) 2 + 1 3 x 3 - 1 ( x + 1 ) 3 + ⋯ ⋯ = log x