Question

4. Show that for any p > 1. This shows that a p-series converges for any p > 1 р— 1 1 to an improper integral Hint: Compare Notice that the sum you need to compare n=2 2, not n 1. Find a function f(x) such that on the interval n - 1,n]. < f(x) starts at n = You might want to draw a picture to help visualize it. Think about right-hand Riemann sums with Ax= 1.

Answer 1

sum of the area of the rectangle lessthanorequalto the arc of under the curve y = f (x) and above x axis and bounde by x = 1 i.e sigma^infinity_n = 2 1/n^e lessthanrequalto integral^infinity_1 1/x^p dx = lim_t rightarrow infinity integral^t_1 x^-p dx = lim_t rightarrow infinity [x^- p + 1/-p + 1]^t_1 = lim_t rightarrow infinity [t^-p + 1/- p + 1 + 1/p - 1] = lim_t rightarrow infinity 1/(1 - p) middot t^p -1 + 1/p -1 = 0 + 1/p -1 [because p > 1] 1/p - 1 so, 1 + signma^infinity_n = 2 1/n^p lessthanorequalto 1 + 1/p - 1 it , signma^infinity_n = 2 1/n^p lessthanorequalto p/p - 1 (proved)