text: # Chapter 11 - Infinite Sequences and Series - 11.3 The Integral Test and Estimates of Sums - 11.3 Exercises - Page 766: 10
  
  Divergent
  
  #### Work Step by Step
  
  The p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$. Given: $\sum_{n=3}^{\infty}n^{-0.9999}=\sum_{n=3}^{\infty}\frac{1}{n^{0.9999}}$ The series $\sum_{n=1}^{\infty}\frac{1}{n^{0.9999}}$is a p-series with $p= 0.9999 \leq 1$ and it is divergent. But the given series is not exactly a p-series, it is because the summation starts from $n=3$ But adding or subtracting finite number of terms will not change the convergence/divergence nature of the series. Hence, the given series is divergent.
  
  After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
metadata: {"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": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}