proof of limit comparison test

The main theorem we will use is the comparison test, which basically states that if $a_{n}>0$ , $b_{n}>0$ and there is an $N$ such that for all $n>N$ , $a_{n}<b_{n}$ , then if $\\sum_{i=1}^{\\infty}b_{n}$ converges so will $\\sum_{i=1}^{\\infty}a_{n}$ .

Suppose $\\lim_{n\\to\\infty}\\frac{a_{n}}{b_{n}}=L$ where $L$ can be a non negative real number or $+\\infty$ .

By definition, for $L$ finite, this means that for every $\\epsilon>0$ there is a natural number $n_{\\epsilon}$ such that for all $n>n_{\\epsilon}$ , $\\left\\|\\frac{a_{n}}{b_{n}}-L\\right\\|<\\epsilon$

To make matters more concrete choose $\\epsilon=\\frac{L}{2}$ and assume $L\eq 0$ and finite.

$0<a_{n}<\\frac{3L}{2}b_{n}$ , for all $n>n_{\\frac{L}{2}}$ .

If $\\sum_{i=1}^{\\infty}b_{n}$ converges, so will $\\sum_{i=1}^{\\infty}\\frac{3L}{2}b_{n}$ and thus by the comparison test, $\\sum_{i=1}^{\\infty}a_{n}$ will also be convergent.

For the reverse result, consider $\\lim_{n\\to\\infty}\\frac{b_{n}}{a_{n}}=\\frac{1}{L}$ , since if $L$ is finite so will $\\frac{1}{L}$ , applying the previous result we can say that if $\\sum_{i=1}^{\\infty}a_{n}$ converges so will $\\sum_{i=1}^{\\infty}b_{n}$

Consider the case $L=0$ , clearly $L=0^{+}$ since both $a_{n}$ and $b_{n}$ are positive, this means that for all $\\epsilon>0$ there exists $n_{\\epsilon}$ such that for all $n>n_{\\epsilon}$ , $0<a_{n}<\\epsilon b_{n}$ .

Considering $\\epsilon=1$ we get the exact formulation of the comparison test, so if $\\sum_{i=1}^{\\infty}b_{n}$ converges so will $\\sum_{i=1}^{\\infty}a_{n}$ .

For the case $L=+\\infty$ just apply the result to $\\lim_{n\\to\\infty}\\frac{b_{n}}{a_{n}}=0$ to conclude that if $\\sum_{i=1}^{\\infty}a_{n}$ converges so will $\\sum_{i=1}^{\\infty}b_{n}$

单词	ProofOfLimitComparisonTest
释义	proof of limit comparison test The main theorem we will use is the comparison test, which basically states that if $a_{n}>0$ , $b_{n}>0$ and there is an $N$ such that for all $n>N$ , $a_{n}<b_{n}$ , then if $\\sum_{i=1}^{\\infty}b_{n}$ converges so will $\\sum_{i=1}^{\\infty}a_{n}$ . Suppose $\\lim_{n\\to\\infty}\\frac{a_{n}}{b_{n}}=L$ where $L$ can be a non negative real number or $+\\infty$ . By definition, for $L$ finite, this means that for every $\\epsilon>0$ there is a natural number $n_{\\epsilon}$ such that for all $n>n_{\\epsilon}$ , $\\left\\\|\\frac{a_{n}}{b_{n}}-L\\right\\\|<\\epsilon$ To make matters more concrete choose $\\epsilon=\\frac{L}{2}$ and assume $L\eq 0$ and finite. $0<a_{n}<\\frac{3L}{2}b_{n}$ , for all $n>n_{\\frac{L}{2}}$ . If $\\sum_{i=1}^{\\infty}b_{n}$ converges, so will $\\sum_{i=1}^{\\infty}\\frac{3L}{2}b_{n}$ and thus by the comparison test, $\\sum_{i=1}^{\\infty}a_{n}$ will also be convergent. For the reverse result, consider $\\lim_{n\\to\\infty}\\frac{b_{n}}{a_{n}}=\\frac{1}{L}$ , since if $L$ is finite so will $\\frac{1}{L}$ , applying the previous result we can say that if $\\sum_{i=1}^{\\infty}a_{n}$ converges so will $\\sum_{i=1}^{\\infty}b_{n}$ Consider the case $L=0$ , clearly $L=0^{+}$ since both $a_{n}$ and $b_{n}$ are positive, this means that for all $\\epsilon>0$ there exists $n_{\\epsilon}$ such that for all $n>n_{\\epsilon}$ , $0<a_{n}<\\epsilon b_{n}$ . Considering $\\epsilon=1$ we get the exact formulation of the comparison test, so if $\\sum_{i=1}^{\\infty}b_{n}$ converges so will $\\sum_{i=1}^{\\infty}a_{n}$ . For the case $L=+\\infty$ just apply the result to $\\lim_{n\\to\\infty}\\frac{b_{n}}{a_{n}}=0$ to conclude that if $\\sum_{i=1}^{\\infty}a_{n}$ converges so will $\\sum_{i=1}^{\\infty}b_{n}$
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