# Tips on studying the convergence of this series

Can someone please offer any tips on how to manipulate the series below, or what convergence test I should use to determine its convergence based on $\alpha$?

$$\sum_{n\ge1} \frac{\sqrt[3]{n+1}+\sqrt[3]{n}}{n^\alpha},\ \alpha \in \mathbb{R}$$

-

## 1 Answer

Hint: It is easy to see that the numerator is $\gt 2n^{1/3}$. And since $n+1\le 2n$, the numerator is $\le (1+2^{1/3})n^{1/3}$. These inequalities will be sufficient for comparison tests.

We really don't need such tight bounds. Any positive constants $a$ and $b$ such that $an^{1/3}\le (n+1)^{1/3}+n^{1/3}\le bn^{1/3}$ are good enough for the comparisons.

Alternately, you could divide top and bottom by $n^{1/3}$. On top you get $\sqrt[3]{1+\frac{1}{n}}+1$, and at the bottom you get $n^{\alpha-1/3}$.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9429324865341187, "perplexity_flag": "head"}