Learning Materials
Features
Discover
Chapter 7: Problem 46
Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n^{3}+1}} $$
Over 30 million students worldwide already upgrade their learning with 91影视!
All the tools & learning materials you need for study success - in one app.
Get started for free
The Fibonacci sequence is defined recursively by \(a_{n+2}=a_{n}+a_{n+1},\) where \(a_{1}=1\) and \(a_{2}=1\) (a) Show that \(\frac{1}{a_{n+1} a_{n+3}}=\frac{1}{a_{n+1} a_{n+2}}-\frac{1}{a_{n+2} a_{n+3}}\). (b) Show that \(\sum_{n=0}^{\infty} \frac{1}{a_{n+1} a_{n+3}}=1\).
In Exercises 85 and \(86,\) (a) find the common ratio of the geometric series, \((b)\) write the function that gives the sum of the series, and (c) use a graphing utility to graph the function and the partial sums \(S_{3}\) and \(S_{5} .\) What do you notice? $$ 1+x+x^{2}+x^{3}+\cdots $$
Show that the series \(\sum_{n=1}^{\infty} a_{n}\) can be written in the telescoping form \(\sum_{n=1}^{\infty}\left[\left(c-S_{n-1}\right)-\left(c-S_{n}\right)\right]\) where \(S_{0}=0\) and \(S_{n}\) is the \(n\) th partial sum.
Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{3 n-1}{2 n+1} $$
Find the sum of the convergent series. $$ \sum_{n=1}^{\infty} \frac{1}{9 n^{2}+3 n-2} $$
What do you think about this solution?
We value your feedback to improve our textbook solutions.
Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine