Section 11.1: Sequences: \( \{ a_n \}_{n=1}^{\infty} \)     Lecture Video

1. Find a formula for the general term \(a_{n}\) of the sequence, assuming that the pattern of the first few terms continues. (Assume that \(n\) begins with 1.)   solution

(a) \(\{2,5,8,11,14, \ldots\}\)   (b) \( \{3, -6, 12, -24, \ldots \} \)   (c) \(\{8, -4, 2, -1, \ldots \}\)   (d) \( \left\{ 2, -\frac{4}{3}, \frac{8}{9},-\frac{16}{27}, \frac{32}{81}, \ldots \right\}\)

2. Determine if the sequence converges or diverges. If it converges, determine its limit.    solution

(a) \( \displaystyle a_n= \frac{2n^2+2n+10}{3n^3 -n^2-5}\)     (b) \( \displaystyle a_n= \frac{2n^4 -3n}{24n^4+ 100n^2+10}\)    (c) \( \displaystyle a_n=\frac{3n^3+8n^2-10}{500n^2+20n+30}\)    (d) \( \displaystyle a_n= 5^n\)    (e) \( \displaystyle a_n= 5 \cdot \left(\frac{2}{3}\right)^n\)  (f) \( \displaystyle a_n= \left( \frac{3}{2}\right)^n\)     (g) \( \displaystyle a_n=3^{-n}\)

3. Determine if the sequence converges or diverges. If it converges, determine its limit.    solution

(h) \(\displaystyle a_n= 4^{\frac{n}{2n+1}}\)     (i) \( \displaystyle a_n = \left(\frac{2n}{3n+1}\right)^2\)    (j) \( \displaystyle a_n= 3^{\frac{n}{n^2+1}}\)   

4. Determine if the sequence converges or diverges. If it converges, determine its limit.  solution

(k) \( \left\{ 2, -\frac{4}{3}, \frac{8}{9},-\frac{16}{27}, \frac{32}{81}, \ldots \right\}\)   (l) \(\{-4, 12, -36, 108, - \ldots \} \)    (m) \(a_n=(-1)^n\)   (n) \( \displaystyle a_n=\frac{(-1)^{n+1}}{n}\)

5. Determine if the sequence converges or diverges. If it converges, determine its limit. solution

(o) \( \displaystyle a_{n}=e^{-\frac{3}{\sqrt n}}\)   (p) \( \displaystyle a_{n}=\frac{(-1)^{n}}{5 \sqrt{n}}\)   (q) \( \displaystyle a_{n}=\frac{n^{2}}{\sqrt{n^{3}+7 n}} \)   (r) \( \displaystyle a_{n}= n^{2}\, e^{-n} \)   (s) \( \displaystyle a_{n}=\ln(3n^2+5)-\ln(n^2+1) \)

6. Determine if the sequence converges or diverges. If it converges, determine its limit.  solution

(t) \(\displaystyle a_n= \frac{5n^2+2n+20}{3n^5 -3n^2-5} \)   (u) \( \displaystyle a_n= \frac{3n^4 -2n}{8n^4+ 100n^2+1000}\)   (v) \( \displaystyle a_n=\frac{2n^3+1}{1500n^2+100n+3000}\)

7. Determine if the sequence converges or diverges. If it converges, determine its limit.   solution

(w) \( \displaystyle a_n= 2^{-n} \)   (x) \( \displaystyle a_n= 3^{n}\, 2^{1-n} \)   (y) \(\displaystyle a_n=4^{n} \)   (z) \( \displaystyle a_n= 2^{\frac{2n}{n+1}} \)