百度试题 结果1 题目determine whether each series converges or diverges.∑limits _(n=1)^(∞ ) (n+1)(n(n+2)) 相关知识点: 试题来源: 解析 the series ∑limits (n+1)(n(n+2)) diverges.反馈 收藏
结果1 题目 determine whether each series converges absolutely, converges conditionally, or diverges.∑limits _(n=1)^(∞ )(-1)^(n-1)(n+1)n 相关知识点: 试题来源: 解析 ∑limits _(n=1)^(∞ )(-1)^(n-1)(n+1)n diverges. 反馈 收藏 ...
Let's say we have a sequence of numbers that is expressed by the term {eq}\displaystyle a_n {/eq}. The sequence is said to be convergent if the terms of the sequence converge to a finite real value as we keep on increasing the value of {eq}\displa...
Determine whether the sequence converges or diverges: a(n)=\frac{4^n}{7^{n+1 Determine whether the sequence converges or diverges. sum_n=1 ^infty 1 sqrtn^3+2 Determine whether the sequence converges and diverges A_n = 2 - (.6)^n. Determine whether the sequence converg...
b. {eq}a_n = \frac {n2^n}{3n} {/eq} Sequence The basic characteristics of a series is whether it converges or diverges. To find it we need to check whether the value of the term keeps on increasing or converges to a finite value as we increase ...
题目 Determine whether the sequence converges or diverges. If it converges, find the limit. a_n=cos (2n) 相关知识点: 试题来源: 解析a_n=cos (2n). As n→ ∞, 2n→ 0, so cos (2n)→ cos 0=1 because cos is continuous. Converges....
解析 converges n(2n^3+1)< n(2n^3)= 1(2n^2)< 1(n^2) for all n≥ 1, so ∑limits_(n=1)^(∞ ) n(2n^3+1) converges by comparison with ∑limits _(n=1)^(∞ ) 1(n^2), which converges because it is a p-series with p=2>1....
n=1}^{\infty} b_n {/eq} converges and {eq}0 \le a_n \le b_n {/eq} for all sufficiently large {eq}n {/eq} (that is, for all {eq}n>N {/eq} for some positive integer {eq}N {/eq}), then the infinite series {eq}\sum_{n=1}^{...
Determine whether the series converges: {eq}\sum\limits_{n = 2}^\infty {{1 \over {n{{\left( {\ln \left( n \right)} \right)}^2}}} {/eq} Convergence and Divergence of Series: A series is said to converge if the limit of partial sum exists...
Determine whether the series converges or diverges: {eq}\; \sum_{k = 1}^{\infty} \frac{(-1)^k}{\sqrt{k} \, + \, 1} {/eq}. Series Test: The series that is given in the question can be either converging or conditionally converging. So to check that we ha...