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Given $u_{n}=\frac{1}{(1+\sqrt{n})^{n}}$ defined for $n \in \mathbb{N}$, what is the nature of $\sum u_{n}$ ?

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Q: 7:53 5G 88 wooclap Got it! Confused Given $u_{n}=\frac{1}{(1+\sqrt{n})^{n}}$ defined for $n \in \mathbb{N}$, what is the nature of $\sum u_{n}$ ? Converging Diverging Submit app.wooclap.com

∻Solution by Steps∻ ‖ step 1 ⋮ We start with the sequence $ u_n = \frac{1}{(1+\sqrt{n})^n} $. To analyze the convergence of the series $ \sum u_n $, we will examine the limit of $ u_n $ as $ n $ approaches infinity. ‖ ‖ step 2 ⋮ We can rewrite $ u_n $ as $ u_n = (1+\sqrt{n})^{-n} $. Taking the limit, we have $ \lim_{n \to \infty} u_n = \lim_{n \to \infty} (1+\sqrt{n})^{-n} $. ‖ ‖ step 3 ⋮ As $ n $ approaches infinity, $ \sqrt{n} $ dominates, and we can approximate $ (1+\sqrt{n})^{-n} $ as $ \left(\sqrt{n}\right)^{-n} = n^{-n/2} $. Thus, $ \lim_{n \to \infty} u_n = 0 $. ‖ ‖ step 4 ⋮ Since $ u_n $ approaches 0, we can apply the ratio test or comparison test to determine the nature of the series. The terms $ u_n $ decrease rapidly, indicating convergence. ‖ ‖ step 5 ⋮ Therefore, the series $ \sum u_n $ converges. ‖ ***A*** ∻Key Concept∻ ⚹Convergence of Series⚹ ∻Explanation∻ ⚹A series converges if the terms approach zero and decrease sufficiently fast, which is the case for $ u_n $.⚹◊What test can be used to determine the convergence or divergence of the series $\sum_{n=1}^{\infty} u_n$ where $u_n = \frac{1}{(1+\sqrt{n})^n}$?⍭ Generate me a similar question◊

Question

7:53
5G
88
wooclap
Got it!
Confused

Given $u_{n}=\frac{1}{(1+\sqrt{n})^{n}}$ defined for $n \in \mathbb{N}$, what is the nature of $\sum u_{n}$ ?

Converging

Diverging
Submit
app.wooclap.com

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ We start with the sequence $ u_n = \frac{1}{(1+\sqrt{n})^n} $. To analyze the convergence of the series $ \sum u_n $, we will examine the limit of $ u_n $ as $ n $ approaches infinity. ‖ ‖ step 2 ⋮ We can rewrite $ u_n $ as $ u_n = (1+\sqrt{n})^{-n} $. Taking the limit, we have $ \lim_{n \to \infty} u_n = \lim_{n \to \infty} (1+\sqrt{n})^{-n} $. ‖ ‖ step 3 ⋮ As $ n $ approaches infinity, $ \sqrt{n} $ dominates, and we can approximate $ (1+\sqrt{n})^{-n} $ as $ \left(\sqrt{n}\right)^{-n} = n^{-n/2} $. Thus, $ \lim_{n \to \infty} u_n = 0 $. ‖ ‖ step 4 ⋮ Since $ u_n $ approaches 0, we can apply the ratio test or comparison test to determine the nature of the series. The terms $ u_n $ decrease rapidly, indicating convergence. ‖ ‖ step 5 ⋮ Therefore, the series $ \sum u_n $ converges. ‖ ***A*** ∻Key Concept∻ ⚹Convergence of Series⚹ ∻Explanation∻ ⚹A series converges if the terms approach zero and decrease sufficiently fast, which is the case for $ u_n $.⚹◊What test can be used to determine the convergence or divergence of the series $\sum_{n=1}^{\infty} u_n$ where $u_n = \frac{1}{(1+\sqrt{n})^n}$?⍭ Generate me a similar question◊

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