CALC II Question on TI-83?

Question

I have just begun infinite series and I am trying to calculate certain limits. Is there a way to calculate this on the TI-83 to check my answer?

For Instance the
(oo=infiniti)

oo
Sigma 3/[n(n+3)]
n=1

=11/6
Is there a way to work this on the TI-83? any help would be appreciated.

Answer 1

There is a way to do series on the TI-83 Plus, so there probably is a way for the TI-83 as well. There is no infinity symbol, however, so you'll have to just put in some large number as the upper limit. You'd check the problem as follows:
sum(seq(3/(x*(x+3)),x,1,500))
I used 500 because it gave an answer fairly close to the value 11/6. The sum() function can be found on my calculator's menus this way: 2nd->STAT(LIST)->MATH->5
seq() can be found this way:
2nd->STAT(LIST)->OPS->5
The sum function takes as its argument a list, and returns the sum of the elements of that list.
The seq function takes as its arguments an expression, the variable in the expression (single variable only, apparently), the start value, and the end value, and it returns a list representing that series.
The result of the above calculation was 1.827357222, and the value of 11/6 is 1.8333..., so I'd say you nailed the question. The calculation took about 15 seconds on my calculator, so I suppose it will take a bit longer on yours. Good luck!

Edit: Oh, yeah . . . if you have to take any more calculus-based courses, and your professors will allow it, I recommend you get a more powerful calculator. A TI Voyage will work wonders for all sorts of problems. It has a built-in computer algebra system and a ton of other features. If you think you might be able to use them, I would encourage you to buy one.

Answer 2

Certainly; you could write a program to do this for you, or you can use the summation function and list functions to produce an approximate answer (remember: integrals are aren't evaluated exactly, either.)

For a program:

:Input Y1
:10000000[STO->]U
:1[STO->]L
:0[STO->]S
:For(A,L,U)
:Y1(A)+S[STO->]S
:End
:Disp S

When the program asks you for Y1 (use the Vars-Y1, not "Y1" as letter and number), enter "3/(n(n+3))" with the quotes.

U is the upperbound. Remember, this is not exact, and if the series converges, the sum will be "close enough" if you set a very high number.
L is the lowerbound.
S is the sum.

Your manual also tells you that the seq() function (See the catalog) allows you to work with series and sequences. Using seq() to generate the sequence:

seq(3/(n(n+3)),n,1, 999)

And the series:

sum(seq(3/(n(n+3)),n,1, 999))



The prototype in pseudocode for seq():

List seq(Expression ex, Variable v, int begin, int end, [int increment])

The prototype in pseudocode for sum:

Number sum(List l[, int start, int end])

It's probably not a good idea to use the seq() and sum() method, though, because you'll need a lot of RAM and it will take a long time. The number 999 can be replaced with any number less than 999, but since the TI83 series cannot generate a list larger than 999 items, it's less exact than the first method.

Answer 3

do not vote for me, this is copied from tetris otaku... it would be very slow to go for(A,1,10000000
You could make the prgm faster with...

:ClrHome
:Input Y1
:DelVar(S
:For(A,1,1000
:Y1(A)+S[STO->]S
:End
:ClrHome
:Disp S

Answer 4

way too many numbers for me

Answer 5

cant get the ques asked

sam ;)