Question

2 $\sum_{k=2}^{\infty} \frac{1}{k(\ln k)^2}$ a) $f(x) = \frac{1}{x(\ln x)^2}$ $u = \ln(x)$ $du = \frac{1}{x}dx$ $\int_2^{\infty} f(x)dx = \int_{\ln 2}^{\infty} \frac{1}{u^2} du = \int_{\ln 2}^{\infty} u^{-2} du$ $= \lim_{t \to \infty} [\frac{-1}{u}]_{\ln 2}^t = \lim_{t \to \infty} [-\frac{1}{t} + \frac{1}{\ln 2}] = \frac{1}{\ln 2}$ $\cdots$ The series converges, because it has b) $a_n = \frac{1}{n(\ln n)^2}$ $a_{n+1} = \frac{1}{(n+1)(\ln(n+1))^2}$ $(n+1)(\ln(n+1))^2 > 1000 \implies n+1 = 60$

          2
$\sum_{k=2}^{\infty} \frac{1}{k(\ln k)^2}$
a) $f(x) = \frac{1}{x(\ln x)^2}$  $u = \ln(x)$ $du = \frac{1}{x}dx$
$\int_2^{\infty} f(x)dx = \int_{\ln 2}^{\infty} \frac{1}{u^2} du = \int_{\ln 2}^{\infty} u^{-2} du$
$= \lim_{t \to \infty} [\frac{-1}{u}]_{\ln 2}^t = \lim_{t \to \infty} [-\frac{1}{t} + \frac{1}{\ln 2}] = \frac{1}{\ln 2}$
$\cdots$ The series converges, because it has
b) $a_n = \frac{1}{n(\ln n)^2}$  $a_{n+1} = \frac{1}{(n+1)(\ln(n+1))^2}$
$(n+1)(\ln(n+1))^2 > 1000 \implies n+1 = 60$

2
∑k=2^∞(1)/(k(ln k)^2)
a) f(x) = (1)/(x(ln x)^2) u = ln(x) du = (1)/(x)dx
∫2^∞ f(x)dx = ∫ln 2^∞(1)/(u^2) du = ∫ln 2^∞ u^-2 du
= limt →∞ [(-1)/(u)]ln 2^t = limt →∞ [-(1)/(t) + (1)/(ln 2)] = (1)/(ln 2)
⋯ The series converges, because it has
b) an = (1)/(n(ln n)^2) an+1 = (1)/((n+1)(ln(n+1))^2)
(n+1)(ln(n+1))^2 > 1000 n+1 = 60

Added by Domingo H.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

11/26/2023

Step 1

Step 1: Start with the inequality (n+1)(ln(n+1))^2 > 1000. Show more…

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(2) sum_{k=2}^{infty} frac{1}{k(ln k)^{2}} a :. The series converges, because it has b I understand the first part, but please help me understand the simplification of (n+1)(ln n +1)^2>1000 to n+1=60 KCenK al 1=(*J x(ln x) u=ln(xau= b)an=! n(ln n)^2 ant= C(n+1)(ln(n) (n+ln(n+1))1000 n+1=66