Respond to this Question
Similar Questions

Calculus
Use the ratio test to find whether the series diverges or converges. 1/5^n (1 to infinity) I think the limit converges to 1/5, so the series converges. 
Calculus
Determine the following about the series. Indicate the test that was used and justify your answer. Sigma (lower index n = 1; upper index infinity) [sin((2n1)pi/2)]/n A. The series diverges B. The series converges conditionally. 
Calculus
For what values of p>0 does the series Riemann Sum [n=1 to infinity] 1/ [n(ln n) (ln(ln n))^p] converge and for what values does it diverge? You need to let the summation start at n = 3 to avoid the singularity at n = 1 (although 
Calculus
The problem with these two questions is that I cannot determine the a and r. The 3rd questionI don't know what I did wrong. Thanks for the help! Tell whether the series converges or diverges. If it converges, give its sum. 
Calculus
Determine whether the series from 0 to infinity of cos(nπ)/(n + 3) converges conditionally or absolutely. A. The series diverges. B. The series converges conditionally but not absolutely. C. The series converges absolutely but 
Calc II
Use the comparison or limit comparison test to decide if the following series converge. Series from n=1 to infinity of (4sin n) / ((n^2)+1) and the series from n=1 to infinity of (4sin n) / ((2^n) +1). For each series which 
Mathematics
State whether this infinite series converges or diverges? 1+(t) + (t^3)+...... t=[(5x+6)/(3x2)] My thoughts on the question: The sum of 'n' terms in a geometric progression is a[r^n  1]/(r1)(let's calk this 1), where r>1 and 
Calc
Does 1/ln(x+1) converge or diverge? I've tried the nth term test, limit comparison test, and integral test. All I get is inconclusive. The other tests I have (geometric series, pseries, telescoping series, alternating series, and 
calculus
determine whether the series converges of diverges the sum from n=1 to infinity of 1/(the square root of (n^3+1)) I said that through the comparision test (comparing to 1/the square root of (n^3) the series converges is this true? 
calculus
determine whether the series converges of diverges the sum from n=1 to infinity of 1/(the square root of (n^3+1)) I said that through the comparision test (comparing to 1/the square root of (n^3) the series converges 
calculus
determine whether the series converges of diverges the sum from k=2 to infinity of (the square root of (ln(k)))/k I said that because you can't integrate the series (goes to infinity) it diverges is this true?