Answer to: Does the series diverge or converge. \sum_{k=1}^{\infty} \frac{g^k}{10^{k-3}} By signing up, you'll get thousands of step-by-step...
Determine the series is converge or diverge: Summation_{n=1}^{infinity} 1 + 3n^2 + n^3/4n^3 - 5n - 2 Determine the series is converge or diverge: Summation_{n=0}^{infinity} (-1)^{n+1} (3^n + 4)/2^n Determine whether the given series converge or dive...
Does the series n = 1 c o s n n n converge absolutely, converge conditionally, or diverge? Give a reason for your answer. Does the series converge or diverge? Give reasons for your answer. \sum_{n=1}^{\infty} e^{-n} Does the series a_n defined by the formu...
Determine whether the following series converge or diverge using the Comparison Test or Limit Comparison Test.∑_(n=1)^∞(1)/(√[n](n+1)(n-1)) 相关知识点: 试题来源: 解析 1/(√[3](n(n+1)(n-1)))=1/(3n^(n-n))1/(√[3](n^3)) 1 7 √[3](n(n+1)(n-1))1/n ...
(sqrt(n) cos^2(n))/((n^2)-2)cos^2(n)总是小于1的 所以分子总是小于sqrt(n)的。分母则是n的平方量级，那个-2在n比较大时可忽略不计。所以整个式子的量级是n的-3/2次幂，比调和级数小多了，converge.
To know how to find the sum of a series in geometric progression, we can use either the finite sum formula or the infinite sum calculation. A geometric series can converge or diverge depending on the value of thecommon ratiorrr. To decide on the convergence vs. divergence of a geometric ...
Does the series 1+ 1(2^3)+ 1(3^3)+⋯ + 1(n^3)+⋯ converge or diverge? 相关知识点: 试题来源: 解析 converges by the p-Series Test.结果一 题目 Does the series converge or diverge? 答案 converges 结果二 题目 Does the series converge or diverge? 答案 Converges because p...
Is ratio test the best test to use? Show with proof. Does the series {eq}\sum_{n = 1}^{\infty} (-1)^n \frac{(2n)!}{2^n n!n} {/eq} converge absolutely, converge conditionally, or diverge?Ratio Test:The ra...
If the sums do not converge, the series is said to diverge.It can go to +infinity, −infinity or just go up and down without settling on any value.Example: 1 + 2 + 3 + 4 + ... Adds up like this: Term Sum so far 1 1 2 3 3 6 4 10 5 15 ... ... The sums are ...
The sum to infinity only exists if -1<r<1. If the common ratio is outside of this range, then the series will diverge and the sum to infinity will not exist. If |r|<1, the sequence will converge to the sum to infinity given by S∞=a/(1-r). ...