l i m n → ∞ [ 1 2 n − 1 + 1 4 n − 2 2 + 1 6 n − 3 2 + ••• + 1 n ] = l i m n → ∞ ∑ k = 1 n [ 1 2 k n − k 2 ] = l i m n → ∞ ∑ k = 1 n 1 n 1 2 ( k n ) − ( k n ) 2 By integral test, ∫ 0 1 1 2 x − x 2 d x = ∫ 0 1 1 1 − ( x − 1 ) 2 d x P u t x − 1 = t ⟹ d x = d t ∫ − 1 0 1 1 − t 2 d t = s i n − 1 t ] − 1 0 = s i n − 1 ( 0 ) − s i n − 1 ( − 1 ) = 0 − ( − π 2 ) = π 2 lim _{n→∞} [ \frac{1}{\sqrt{2n-1}}+\frac{1}{\sqrt{4n-2^2}}+\frac{1}{\sqrt{6n-3^2}}+•••+\frac{1}{n}]=lim _{n→∞}\sum_{k=1 } ^n [ \frac{1}{\sqrt{2kn-k^2}}]\\
=lim _{n→∞}\sum_{k=1 } ^n \frac{1}{n}
\frac{1}{\sqrt{2(\frac{k}{n})-(\frac{k}{n})
^2}}\\
\text{By integral test,}\\
\int_0^1 \frac{1}{\sqrt{2x-x^2}}dx\\
=\int_0^1 \frac{1}{\sqrt{1-(x-1)^2}}dx\\
Put \space x-1=t\implies dx=dt\\
\int_{-1}^0 \frac{1}{\sqrt{1-t^2}}dt\\
=sin^{-1}t]_{-1}^0\\
=sin^{-1}(0)-sin^{-1}(-1)\\
=0-(-\frac{\pi}{2})\\
=\frac{\pi}{2} l i m n → ∞ [ 2 n − 1 1 + 4 n − 2 2 1 + 6 n − 3 2 1 + ••• + n 1 ] = l i m n → ∞ ∑ k = 1 n [ 2 kn − k 2 1 ] = l i m n → ∞ ∑ k = 1 n n 1 2 ( n k ) − ( n k ) 2 1 By integral test, ∫ 0 1 2 x − x 2 1 d x = ∫ 0 1 1 − ( x − 1 ) 2 1 d x P u t x − 1 = t ⟹ d x = d t ∫ − 1 0 1 − t 2 1 d t = s i n − 1 t ] − 1 0 = s i n − 1 ( 0 ) − s i n − 1 ( − 1 ) = 0 − ( − 2 π ) = 2 π
