Prove: $sum b_n < infty Longrightarrow sum a_n < infty $, where $ exists N: forall n geq N:...
Given two positive sums: $sum_{n=1}^{infty} a_n $ and $ sum_{n=1}^{infty} b_n $.
$ exists N: forall n geq N: frac{a_{n+1}}{a_n} leq frac{b_{n+1}}{b_n}$
Prove: $sum {b_n} < infty Longrightarrow sum a_n < infity $
I tried to use the ratio test, but I can't assume that $ frac{b_{n+1}}{b_n} leq b_n$.
Even though I know that $lim _{n to infty} b_n = 0$, that's not enough to determine something about $ frac{b_{n+1}}{b_n}$.
Especially because the Ratio test is not of the form of "if and only if", so How can I use the fact that $ frac{a_{n+1}}{a_n} leq frac{b_{n+1}}{b_n}$ in order to determine a fact about $sum {a_n} $ convergence?
real-analysis calculus convergence
asked Dec 8 at 15:29
Jneven
746320
add a comment |
Given two positive sums: $sum_{n=1}^{infty} a_n $ and $ sum_{n=1}^{infty} b_n $.
$ exists N: forall n geq N: frac{a_{n+1}}{a_n} leq frac{b_{n+1}}{b_n}$
Prove: $sum {b_n} < infty Longrightarrow sum a_n < infity $
I tried to use the ratio test, but I can't assume that $ frac{b_{n+1}}{b_n} leq b_n$.
Even though I know that $lim _{n to infty} b_n = 0$, that's not enough to determine something about $ frac{b_{n+1}}{b_n}$.
Especially because the Ratio test is not of the form of "if and only if", so How can I use the fact that $ frac{a_{n+1}}{a_n} leq frac{b_{n+1}}{b_n}$ in order to determine a fact about $sum {a_n} $ convergence?
real-analysis calculus convergence
asked Dec 8 at 15:29
Jneven
746320
add a comment |
Given two positive sums: $sum_{n=1}^{infty} a_n $ and $ sum_{n=1}^{infty} b_n $.
$ exists N: forall n geq N: frac{a_{n+1}}{a_n} leq frac{b_{n+1}}{b_n}$
Prove: $sum {b_n} < infty Longrightarrow sum a_n < infity $
I tried to use the ratio test, but I can't assume that $ frac{b_{n+1}}{b_n} leq b_n$.
Even though I know that $lim _{n to infty} b_n = 0$, that's not enough to determine something about $ frac{b_{n+1}}{b_n}$.
Especially because the Ratio test is not of the form of "if and only if", so How can I use the fact that $ frac{a_{n+1}}{a_n} leq frac{b_{n+1}}{b_n}$ in order to determine a fact about $sum {a_n} $ convergence?
real-analysis calculus convergence
asked Dec 8 at 15:29
Jneven
746320
Given two positive sums: $sum_{n=1}^{infty} a_n $ and $ sum_{n=1}^{infty} b_n $.
$ exists N: forall n geq N: frac{a_{n+1}}{a_n} leq frac{b_{n+1}}{b_n}$
Prove: $sum {b_n} < infty Longrightarrow sum a_n < infity $
I tried to use the ratio test, but I can't assume that $ frac{b_{n+1}}{b_n} leq b_n$.
Even though I know that $lim _{n to infty} b_n = 0$, that's not enough to determine something about $ frac{b_{n+1}}{b_n}$.
Especially because the Ratio test is not of the form of "if and only if", so How can I use the fact that $ frac{a_{n+1}}{a_n} leq frac{b_{n+1}}{b_n}$ in order to determine a fact about $sum {a_n} $ convergence?
real-analysis calculus convergence
real-analysis calculus convergence
asked Dec 8 at 15:29
Jneven
746320
asked Dec 8 at 15:29
Jneven
746320
asked Dec 8 at 15:29
Jneven
746320
asked Dec 8 at 15:29
Jneven
746320
asked Dec 8 at 15:29
Jneven
746320
746320
add a comment |
add a comment |
1 Answer
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Note that ${frac{a_n}{b_n}}_{nin Bbb N}$ is a decreasing sequence of positives numbers, hence ${frac{a_n}{b_n}}_{nin Bbb N}$ is bounded , therefore there is an $M>0$ with $a_nleq M b_n$ for all $nin Bbb N$. Hence $sum_{n=1}^{infty}a_nleq M sum_{n=1}^{infty}b_n<inftyimplies sum_{n=1}^{infty}a_n<infty.$
answered Dec 8 at 15:58
UserS
1,5371112
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1 Answer
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Note that ${frac{a_n}{b_n}}_{nin Bbb N}$ is a decreasing sequence of positives numbers, hence ${frac{a_n}{b_n}}_{nin Bbb N}$ is bounded , therefore there is an $M>0$ with $a_nleq M b_n$ for all $nin Bbb N$. Hence $sum_{n=1}^{infty}a_nleq M sum_{n=1}^{infty}b_n<inftyimplies sum_{n=1}^{infty}a_n<infty.$
answered Dec 8 at 15:58
UserS
1,5371112
add a comment |
Note that ${frac{a_n}{b_n}}_{nin Bbb N}$ is a decreasing sequence of positives numbers, hence ${frac{a_n}{b_n}}_{nin Bbb N}$ is bounded , therefore there is an $M>0$ with $a_nleq M b_n$ for all $nin Bbb N$. Hence $sum_{n=1}^{infty}a_nleq M sum_{n=1}^{infty}b_n<inftyimplies sum_{n=1}^{infty}a_n<infty.$
answered Dec 8 at 15:58
UserS
1,5371112
add a comment |
Note that ${frac{a_n}{b_n}}_{nin Bbb N}$ is a decreasing sequence of positives numbers, hence ${frac{a_n}{b_n}}_{nin Bbb N}$ is bounded , therefore there is an $M>0$ with $a_nleq M b_n$ for all $nin Bbb N$. Hence $sum_{n=1}^{infty}a_nleq M sum_{n=1}^{infty}b_n<inftyimplies sum_{n=1}^{infty}a_n<infty.$
answered Dec 8 at 15:58
UserS
1,5371112
Note that ${frac{a_n}{b_n}}_{nin Bbb N}$ is a decreasing sequence of positives numbers, hence ${frac{a_n}{b_n}}_{nin Bbb N}$ is bounded , therefore there is an $M>0$ with $a_nleq M b_n$ for all $nin Bbb N$. Hence $sum_{n=1}^{infty}a_nleq M sum_{n=1}^{infty}b_n<inftyimplies sum_{n=1}^{infty}a_n<infty.$
answered Dec 8 at 15:58
UserS
1,5371112
answered Dec 8 at 15:58
UserS
1,5371112
answered Dec 8 at 15:58
UserS
1,5371112
answered Dec 8 at 15:58
UserS
1,5371112
1,5371112
add a comment |
add a comment |
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