Determine whether or not each of the following infinite series converges, and evaluate it if it does.
- [a [math]\sum_{i=0}^\infty \frac7{5^i}[/math]
- [math]\sum_{k=1}^\infty \frac{a}{5^k}[/math]
- [math]\sum_{n=1}^\infty \left( \frac1{2^n} + \frac1n \right)[/math]
- [math]\sum_{j=0}^\infty \left(\frac1{2^j} - \frac1{3^j}\right)[/math]
- [math]\sum_{i=1}^\infty \frac{5\cdot2^i+6i}{i2^i}[/math]
- [math]\sum_{k=0}^\infty \left(3+\frac1{3^k}\right)[/math]
- [math]\sum_{i=1}^\infty \frac{i^2-1}{i^2+1}[/math]
- [math]\sum_{k=0}^\infty \frac{2^k+3^k}{6^k}[/math].
Consider the infinite series [math]\sum_{i=0}^\infty a_i[/math] defined by
Write out the sum of the first ten terms. Does the series converge? If so, to what value?
Using Theorem \ref{thm 9.2.2}, show that if [math]\sum_{i=m}^\infty a_i[/math] converges and if [math]\sum_{i=m}^\infty b_i[/math] diverges, then [math]\sum_{i=m}^\infty (a_i+b_i)[/math] must diverge.
Is it true that if the series [math]\sum_{i=m}^\infty (a_i+b_i)[/math] converges, then both [math]\sum_{i=m}^\infty a_i[/math] and [math]\sum_{i=m}^\infty b_i[/math] must also converge? Give a reason for your answer.
Prove that the harmonic series [math]\sum_{k=1}^\infty \frac1k[/math] diverges using the following elementary argument. Begin by grouping the terms of the series:
and observe that
Consider the infinite series [math]\sum_{k=1}^\infty \left(\frac1k - \frac1{k+1}\right)[/math]. By writing out a few terms of the sequence of partial sums, show that the series converges, and give its value.
An infinite series of the form [math]\sum_{i=m}^\infty (a_i - a_{i+1})[/math] is called a telescoping series (see Problem Exercise). Prove that it converges if and only if the sequence [math]\{ s_n \}[/math] converges. If it does converge, what is its value?
Determine whether or not each of the following infinite series converges, and evaluate it if it does.
- [math]\sum_{k=1}^\infty \frac1{k(k+1)}[/math].
- [math]\sum_{i=1}^\infty \frac{2i+1}{i^2(i^2+2i+1)}[/math].
- [math]\sum_{k=1}^\infty \ln \left(\frac{k+1}k\right) = \ln \left(\frac21\right) + \ln \left(\frac32\right) + \ln \left(\frac43\right) + \cdots [/math].