Calculus Multiple Choice

• Feb 23rd 2006, 03:30 PM
Nimmy
Calculus Multiple Choice
Mutliple Choice questions

5. Determine which converges:

a. $\sum^{\infty}_{n=1}(4+(-1)^n)$.
b. $\sum^{\infty}_{n=1}2^n/(n+1)!$.
c. $\sum^{\infty}_{n=0} 5(3/2)^n$.
d. $\sum^{\infty}_{n=1} 1/n^(1/2)$.
e. None of these

7. Determine which test can be proved the divergence $\sum^{\infty}_{n=1} 1/n^(1/2)$.

a. Geometric Series Test
b. P Series Test
c. Ratio Test
d. Nth Term Test for Divergence
e. None of These

14. Determine which of the following test could be used to show that $\sum^{\infty}_{n=1} (2n-1)/(3n+5)$.

a. Root Test
b. Ratio Test
c. Geometric Series Test
d. P-Series Test
e. None of These

17. Find the radius of convergence of the power series $\sum^{\infty}_{n=0} (x/2)^n$.

a. 1/2
b. 2
c. Infinity
d. 0
e. None of These
• Feb 23rd 2006, 05:11 PM
ThePerfectHacker
Problem 5:
a)This diverges by oscillation.
b)This converges.
c)This is a geometric series with the constant ration $|3/2|>1$ thus it diverges.
d)This is called the p-series, since the exponent is less than 1 it must diverge.

Explaination to b:
Apply the ratio test,
$a_{k+1}=\frac{2^{k+1}}{(k+2)!}$
$a_k=\frac{2^k}{(k+1)!}$
Thus,
$\lim_{k\to\infty}\frac{2^{k+1}}{(k+2)!}\frac{(k+1) !}{2^k}=\frac{2}{k+2}=0<1$
Thus, it converges.
Q.E.D.
• Feb 23rd 2006, 07:52 PM
Nimmy
Quote:

Originally Posted by Nimmy
Mutliple Choice questions

5. Determine which converges:

a. $\sum^{\infty}_{n=1}(4+(-1)^n)$.
b. $\sum^{\infty}_{n=1}2^n/(n+1)!$.
c. $\sum^{\infty}_{n=0} 5(3/2)^n$.
d. $\sum^{\infty}_{n=1} 1/n^(1/2)$.
e. None of these

7. Determine which test can be proved the divergence $\sum^{\infty}_{n=1} 1/n^(1/2)$.

a. Geometric Series Test
b. P Series Test
c. Ratio Test
d. Nth Term Test for Divergence
e. None of These

14. Determine which of the following test could be used to show that $\sum^{\infty}_{n=1} (2n-1)/(3n+5)$.

a. Root Test
b. Ratio Test
c. Geometric Series Test
d. P-Series Test
e. None of These

17. Find the radius of convergence of the power series $\sum^{\infty}_{n=0} (x/2)^n$.

a. 1/2
b. 2
c. Infinity
d. 0
e. None of These

I still need help on 7,14,17.
• Feb 24th 2006, 08:29 AM
ThePerfectHacker
Quote:

Originally Posted by Nimmy
I still need help on 7,14,17.

7)That is the p-series, because it has form $\frac{1}{k^n}$.

14)None of these, you may solve it using a different test. The limit comparison test.

17)Use the ratio test, $a_k=(x/2)^k$ and $a_{k+1}=(x/2)^{k+1}$.
Thus,
$\lim_{k\to\infty}\frac{x^{k+1}}{2^{k+1}}\frac{2^k} {x^k}=\frac{x}{2}$ Now it is necessary for this is converge when $|\frac{x}{2}|<1$ that happens when $-2 thus the radius of convergence is 2.
• Feb 24th 2006, 01:11 PM
Rich B.
Hi Folks:

Are we doing homework or completing take-home tests for students of dimished ambition? ...just a thought (not an insinuation).

Respectfully,

Rich B.