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1 solution
Your Taylor expansion for 1 / ( 1 + x ) is wrong and at the end you have n → ∞ lim k = 1 ∑ ∞ n + k 1 which is infinity and not 0 because k = 1 ∑ ∞ n + k 1 = ∞ .
If fact if you fix your Taylor expansion it works because at the end you have the limit of a convergent series minus its partial sum which gives 0.
I mistakenly viewed the solution. So I'm sharing my idea here in the comment section. We can show that for n > 1 . n → ∞ lim ∫ 0 1 1 + x x n d x ≈ n → ∞ lim ( n 1 − ln ( n n − 1 ) − n − 1 1 ) and hence as n → ∞ the limit of integral goes to zero.
In fact the integral has closed form in terms of Harmonic numbers.
I = ∫ 0 1 x + 1 x n d x = ∫ 0 1 ( x n k = 0 ∑ ∞ ( − x ) k ) d x = k = 0 ∑ ∞ ∫ 0 1 ( − 1 ) k x n + k d x = k = 0 ∑ ∞ n + k + 1 ( − 1 ) k = k = 1 ∑ ∞ n + k ( − 1 ) k For ∣ x ∣ < 1
Therefore, n → ∞ lim I = n → ∞ lim k = 1 ∑ ∞ n + k ( − 1 ) k = 0 .