Math Challenges

Submissions for Problem #49

Problem #49

Explain whether the series is convergent or divergent.

\[ \displaylines{\text{Is }\sum a_{n}\text{ convergent if}\\ 1)\text{ }a_{n}=\sqrt{n+1}-\sqrt{n}\\ 2)\text{ }a_{n}=\frac{\sqrt{n+1}-\sqrt{n}}{n}\\ 3)\text{ }a_{n}=\left(\sqrt[n]{n}-1\right)^{n}\\ 4)\text{ }a_{n}=\frac{1}{1+z^{n}}\text{ for }z\in\mathbb{C}} \]

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zapwai

Solution: \( \displaylines{1)S_{N}=\sqrt{N+1}-1\text{ which is divergent as }N\to\infty\\ 2)\text{Grouping the first 3, then next 5, then next 7 terms, etc. we can bound the partial sum by}\\ S_{N}\leq1\left(2-1\right)+\frac14\left(3-2\right)+\frac19\left(4-3\right)+\cdots=\sum_1^{\infty}\frac{1}{n^2}\\ \text{So the series is convergent by comparison with a convergent p-series.}\\ 3)\text{Root test gives convergence, as }\left|a_{n}\right|^{\frac{1}{n}}=n^{\frac{1}{n}}-1\to0\text{ as }n\to\infty\\ \text{(let }y=n^{\frac{1}{n}}\text{ and take the limit of its logarithm to see that it tends to 1)}\\ 4)\text{For }\left|z\right|\lt1\text{ the series diverges, as the terms }a_{n}=\frac{1}{1+z^{n}}\text{ are greater than or equal to }\frac12.\\ \text{For }\left|z\right|\gt1\text{ the partial sum is bounded by a geometric series with }r=\frac{1}{\left|z\right|}\text{ and is convergent.}} \)