Question

1. Find the general term of the sequence, starting with n=1, determine whether the sequence converges, and if so find its limit.

9/10, 9/20, 29/30, 39/40 ...

2. Find the general term of the sequence, starting with n=1 determine whether the sequence converges, and if so find its limit. The general term of the sequence is bracket a sub n bracket, positive where n=1.

19, 19/7, 19/ 49, 19/343 ...

3. Starting n=1, and considering the even and odd terms separately, find a formula for the general term of the sequence

1, 1/81, 3, 1/9^4, 5, 1/9^6 ...

4. Consider the sequence

a1 = 90^1 /2
a2 = 90^ 1/2 + 90^1/4
a3 = 90^1/2 +90^1/4 +90^1/8

A. find the recursion formula for a sub n+1.
B. Assuming that the sequence converges, find its limit L

Answer 1

1.

$a_n=\frac{9+10(n-1)}{10n}$

$\lim\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}\frac{9+10(n-1)}{10n}=1$

2.

$a_n=\frac{19}{7^{n-1}}$

$\lim\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}\frac{19}{7^{n-1}}=0$

3.

$a_{2n-1}=2n-1; a_{2n}=\frac{1}{9^{2n}}$

$a_n=n|\sin\frac{\pi n}{2}|+\frac{|\cos\frac{\pi n}{2}|}{9^n}$

4.

$a_{n+1}=a_n+90^{\frac{1}{2^{n+1}}}$

$\lim\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}a_{n+1}=\lim\limits_{n\to\infty}(a_n+90^{\frac{1}{2^{n+1}}})= \lim\limits_{n\to\infty}a_n+1$

Hence

$\lim\limits_{n\to\infty}a_n=\infty$