Sequences & Series: Convergence Criteria
Sequences & Series: Convergence Criteria
A sequence $(a_n)$ converges to $L$ if $\forall\epsilon>0,\exists N$ such that $n>N\Rightarrow|a_n-L|<\epsilon$. The Cauchy criterion gives an intrinsic convergence test without knowing the limit: $(a_n)$ converges iff it is a Cauchy sequence. Series $\sum a_n$ are defined as limits of partial sums, and the comparison, ratio, root, and integral tests determine convergence without computing the sum.
Cauchy Sequences & Convergence
A sequence $(a_n)$ in $\mathbb{R}$ is Cauchy if: $\forall\epsilon>0,\exists N$ such that $m,n>N\Rightarrow|a_m-a_n|<\epsilon$. Cauchy criterion: $(a_n)$ converges iff it is Cauchy (completeness of $\mathbb{R}$). Every convergent sequence is Cauchy: $|a_m-a_n|\leq|a_m-L|+|L-a_n|<\epsilon/2+\epsilon/2$. Every Cauchy sequence is bounded. The converse (Cauchy $\Rightarrow$ convergent) requires completeness and fails in $\mathbb{Q}$.
Ratio & Root Tests
Let $\sum a_n$ be a series with $a_n>0$. Ratio test: if $\lim\frac{a_{n+1}}{a_n}=L$, then $\sum a_n$ converges if $L<1$, diverges if $L>1$, inconclusive if $L=1$. Root test: if $\limsup a_n^{1/n}=L$, then converges if $L<1$, diverges if $L>1$. Proof: if $L<1$, pick $r\in(L,1)$ and $N$ so $a_{n+1}/a_n
Example 1
Prove $\sum_{n=1}^\infty 1/n^2$ converges but $\sum 1/n$ diverges.
Solution: $\sum 1/n^2$: use comparison with telescoping. $1/n^2<1/(n(n-1))=1/(n-1)-1/n$, so $\sum_{n=2}^N 1/n^2<\sum_{n=2}^N(1/(n-1)-1/n)=1-1/N<1$. Partial sums are bounded, terms positive — converges. $\sum 1/n$: Cauchy criterion fails (Harmonic series diverges). Group: $\sum_{n=2^k+1}^{2^{k+1}}1/n\geq 2^k\cdot 1/2^{k+1}=1/2$. So partial sums $S_{2^N}>1+N/2\to\infty$.
Example 2
Determine convergence of $\sum_{n=1}^\infty n!/n^n$.
Solution: Ratio test: $\frac{a_{n+1}}{a_n}=\frac{(n+1)!/(n+1)^{n+1}}{n!/n^n}=\frac{(n+1)n^n}{(n+1)^{n+1}}=\frac{n^n}{(n+1)^n}=\left(\frac{n}{n+1}\right)^n=\frac{1}{(1+1/n)^n}\to\frac{1}{e}<1$. By ratio test, the series converges. Note: $\sum n!/n^n=e^{-1}\cdot e\cdot\sum$ (connection to Stirling's approximation).
Practice
- Prove: if $(a_n)$ converges then $|a_n|\to|L|$ but not conversely.
- Show $\sum_{n=0}^\infty x^n=1/(1-x)$ for $|x|<1$ using partial sums and convergence.
- Determine convergence of $\sum(-1)^n/\sqrt{n}$ using the alternating series test.
- Prove the condensation test: for decreasing $a_n\geq 0$, $\sum a_n$ converges iff $\sum 2^n a_{2^n}$ converges.
Show Answer Key
1. If $a_n\to L$ then $|a_n-L|\to0$, so $||a_n|-|L||\le|a_n-L|\to0$ by reverse triangle inequality, giving $|a_n|\to|L|$. Converse fails: $a_n=(-1)^n$ has $|a_n|\to1$ but $(a_n)$ diverges.
2. $S_n=\sum_{k=0}^{n}x^k=\frac{1-x^{n+1}}{1-x}$. For $|x|<1$, $x^{n+1}\to0$, so $S_n\to\frac{1}{1-x}$.
3. $a_n=1/\sqrt{n}$ is decreasing and $a_n\to0$, so $\sum(-1)^n/\sqrt{n}$ converges by the alternating series test. (It converges conditionally, not absolutely.)
4. For decreasing $a_n\ge0$: $2^n a_{2^n}\le\sum_{k=2^{n-1}+1}^{2^n}a_k\le 2^{n-1}a_{2^{n-1}}$ (by monotonicity and counting terms). Summing over $n$ shows $\sum a_n$ and $\sum 2^n a_{2^n}$ converge or diverge together.