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Recitation 19

Example 1: Determine whether the series is convergent or divergent. If it is convergent, find its sum. (a) 1+3n2n\sum\frac{1+3^n}{2^n}; (b) 11+(2/3)n\sum\frac{1}{1+(2/3)^n}; (c) (3/5n+2/n)\sum (3/5^n+2/n).

Hint: (a) (b) Find the limit of ana_n. (c) If an\sum a_n is convergent and bn\sum b_n is divergent, then (an+bn)\sum (a_n + b_n) is divergent.

Example 2: Determine whether the series n=13n(n+3)\sum_{n=1}^{\infty}\frac{3}{n(n+3)} is convergent or divergent by expressing sns_n as a telescoping sum. If it is convergent, find its sum.

Hint: Note that 3k(k+3)=1k1k+3\frac{3}{k(k+3)} = \frac{1}{k} - \frac{1}{k+3}. We have sn=k=1n3k(k+3)=k=1n(1k1k+3)=1/1+1/2+1/31/(n+1)1/(n+2)1/(n+3)s_n = \sum_{k=1}^n\frac{3}{k(k+3)} = \sum_{k=1}^n(\frac{1}{k} - \frac{1}{k+3}) = 1/1 + 1/2 + 1/3 - 1/(n+1) - 1/(n+2) - 1/(n+3).

Example 3: Find the values of xx for which the series n=0(4)n(x5)n\sum_{n=0}^\infty (-4)^n(x-5)^n converges. find the sum of the series for those values of xx.

Hint: This is a geometric series with initial term 11 and common ratio 4(x5)-4(x-5). It is convergent if and only if 1<4(x5)<1-1 < -4(x-5) < 1, i.e., 19/4<x<21/419/4 < x < 21/4. When it converges, the sum is 1/(1(4)(x5))=1/(4x19)1 / (1 - (-4)(x-5)) = 1/(4x-19).

Example 4: Use the integral test to determine whether the series n=1nn2+1\sum_{n=1}^\infty \frac{n}{n^2+1} is convergent or divergent.

Hint: Consider f(x)=xx2+1f(x) = \frac{x}{x^2+1} for x>1x > 1. Check it is continuous, positive and ultimately decreasing.

Example 5: Determine whether the series is convergent or divergent. (a) 1+1/8+1/27+1/64+1/125+1 + 1/8 + 1/27 + 1/64 + 1/125 + \ldots; (b) n=11n2+4\sum_{n=1}^\infty\frac{1}{n^2+4}; (c) n=21nlnn\sum_{n=2}^\infty\frac{1}{n\ln n}.

Hint: Consider f(x)=1/x3f(x) = 1/x^3, f(x)=1x2+4f(x) = \frac{1}{x^2+4}, f(x)=1xlnxf(x) = \frac{1}{x\ln x} for (a), (b), (c) respectively. Check they are continuous, positive and ultimately decreasing.