Re: Convergence of sequences

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**llamaguy** Assume that

lim_{(n->}_{∞)} a_{n} = a and lim_{(n->}_{∞)} b_{n }= b

Show that

I. lim_{(n->}_{∞)} (a_{n} + b_{n}) = a + b

II. lim_{(n->}_{∞)} a_{n}b_{n} = ab

$|a_n+b_n-a-b|\le |a_n-a|+|b_n-b|$ and $|a_nb_b-ab|\le |a_nb_n-b_na|+|ab_n-ab|$

Re: Convergence of sequences

Start with the definition for convergence of sequences. The first is extremely straightforward. The second might require some manipulation.

Re: Convergence of sequences

Quote:

Originally Posted by

**Plato** $|a_n+b_n-a-b|\le |a_n-a|+|b_n-b|$ and $|a_nb_b-ab|\le |a_nb_n-b_na|+|ab_n-ab|$

Appreciate the help, not entirely sure how you got there? Which is which, from the definition given in my book?

**Limit of a sequence**

We say that a sequence ${a_n}$ converges to the limit $L$, and we write $lim_n→∞$, if for every positive real number ε there exists an integer $N$ (which may depend on $ε$) such that if $n≥N$, then $|a_n-L|< ε$

Re: Convergence of sequences

Quote:

Originally Posted by

**llamaguy** Appreciate the help, not entirely sure how you got there? Which is which, from the definition given in my book?

**Limit of a sequence**

We say that a sequence ${a_n}$ converges to the limit $L$, and we write $lim_n→∞$, if for every positive real number ε there exists an integer $N$ (which may depend on $ε$) such that if $n≥N$, then $|a_n-L|< ε$

We here do not give tutorials. Nor do we do homework for you. You show some effort and we help.

The first one is trivial. The second one depends on knowing a convergent sequence is bounded.

$\left( {\exists B > 0} \right)\left( {\forall n} \right)\left[ {\left| {{b_n}} \right| \leqslant B} \right]$

Now use the textbook's definition to get

$\left| {{a_n} - a} \right| < \dfrac{\varepsilon }{{2\left( B \right)}}\,\& \,\left| {{b_n} - b} \right| < \dfrac{\varepsilon }{{2\left( {1 + \left| a \right|} \right)}}$

Re: Convergence of sequences

Quote:

Originally Posted by

**Plato** The first one is trivial. The second one depends on knowing a convergent sequence is bounded.

$\left( {\exists B > 0} \right)\left( {\forall n} \right)\left[ {\left| {{b_n}} \right| \leqslant B} \right]$

For the second one, you can also use

$\displaystyle \begin{align*}\left|a_nb_n - ab\right| & = \left|a_nb_n - (a_nb + ab_n - ab) + (a_nb+ab_n - ab) - ab\right| \\ & = \left|(a_nb_n - a_nb -ab_n + ab) + (a_nb - ab) + (ab_n - ab)\right| \\ & = \left|(a_n - a)(b_n - b) + b(a_n - a) + a(b_n - b)\right| \\ & \le |a_n-a||b_n-b| + |b||a_n-a| + |a||b_n-b|\end{align*}$

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Originally Posted by

**llamaguy** Appreciate the help, not entirely sure how you got there? Which is which, from the definition given in my book?

You should review the triangle inequality. Plato did not give you the solution. He gave you a hint to finding the solution (as did I). Given $\varepsilon>0$, you need to find the correct values for $N$ that will show $|a_N + b_N - a-b| < \varepsilon$ and $M$ so that $|a_Mb_M - ab| < \varepsilon$ yourself.

Re: Convergence of sequences

This isn't homework, I'm learning this independently. The problem is that the tasks I have worked on so far using the definition of a convergence of a series use an $n$ term in the sequence, f.ex.:

show that $ lim(n->∞)|c/n^p|=0$ for any real $c$ and any $p>0$

solution: I set $ε>0$, then

$|c/n^p|<ε$ if $n^p>|c/ε|$ which implies $n>|c/ε|$ raised to $(1/p)$, so if you set $N=|c/ε|$ raised to $(1/p)$ it satisfies the definition.

I just don't know how to apply any of this when the initial sequence doesn't directly contain an $n$ term, and while I'm appreciative of your help I just don't get it, I'm sorry. The textbook doesn't help a single bit and I don't really have anywhere to ask for further help. Not sure if I am allowed to ask, but do you if anything then know a site where I can ask for a full tutorial on this problem? :/

Re: Convergence of sequences

Quote:

Originally Posted by

**llamaguy** This isn't homework, I'm learning this independently. The problem is that the tasks I have worked on so far using the definition of a convergence of a series use an $n$ term in the sequence, f.ex.:

show that $ lim(n->∞)|c/n^p|=0$ for any real $c$ and any $p>0$

solution: I set $ε>0$, then

$|c/n^p|<ε$ if $n^p>|c/ε|$ which implies $n>|c/ε|$ raised to $(1/p)$, so if you set $N=|c/ε|$ raised to $(1/p)$ it satisfies the definition.

I just don't know how to apply any of this when the initial sequence doesn't directly contain an $n$ term, and while I'm appreciative of your help I just don't get it, I'm sorry. The textbook doesn't help a single bit and I don't really have anywhere to ask for further help. Not sure if I am allowed to ask, but do you if anything then know a site where I can ask for a full tutorial on this problem? :/

Here is a good webpage.

It seem to me that you are approaching this backwards.

You must know sequence convergence before you can understand series.

Look at chapter three of that online textbook.

Learn to do proofs about simple properties of sequence convergence.

Re: Convergence of sequences

Thank you, I'll check through that :D