Calculus: Sum of a Sequence!

In the process of trying to prove the convergency of the sum of a sequence--or, more spacifically, the divergency of it--I attempted to change the function that I was working with into something I was familiar with, but in doing so I ended up creating a proof which seems very wrong--mainly because the proof suggests that 0=1, which clearly isn't true. I was hopeing that someone could prove my proof wrong because it is bugging me, and I have no clue why it is wrong.
And since it is wrong (or so I assume it is) can someone actually prove by other means why the particular function is divergent.

Here goes:

I had to prove why the following function is divergent (has no definate value):

Sum (x=1 to infinity) sin(1/x)

I know that this graph looks like a decrasing exponential function from x=1/pi to infinity (for those who don't know, an exponential function is a function that can be written as x^n, where n is a constant and x is a variable; a decreasing geometric series has r: r<0). I know how to prove whether an exponential function is convergent or divergent so I attempted to change the function into one.

n is an arbitrary variable:

sin(1/x)=1/x^n
ln(sin(1/x))=n(ln(1/x))
n=ln(sin(1/x))/ln(1/x)

I believe that if I could determin what the value of n was as x approached infinity, it would help me to determin whether the function was convergent or divergent, so I did just that.

lim (x->infinity) n=?
lim (x->infinity) ln(sin(1/x))/ln(1/x)

this value becomes undefined/infinity, therefore I needed to use the h' pitol (I have no clue what the guy's name is or how to spell it) rule to find the actual limit.

lim (x->infinity) (-1/x^2(cos(1/x)/sin(1/x)))/(-1/x^2/(1/x))
lim (x->infinity) tan(1/x)/x

this becomes 0/infinity, which equals 0; therefore:

lim (x->infinity) n=0

According to this:

lim (x->infinity) 1/x^n = 1/(infinity)^0 = 1
yet:
lim (x->infinity) sin(1/x) = 0

If the two functions are equal than how can the above limits be different? They must be equal because I set them equal to each other and solved for n, yet they are not. Why?

Please explain to me why the above proof is wrong, and please also explain why the summation of sin(1/x) from 1 to infinity is divergent. I also used a rule in my proof, the H' pitol rule (or something like that), what is the actual name of that rule?

P.S. I am a AP Calculus BC student, so if you are able to explain this, please do so in terms that I can understand.

Thanks
Econundrum

Originally posted by econundrum
I know that this graph looks like a decrasing exponential function from x=1/pi to infinity (for those who don't know, an exponential function is a function that can be written as x^n, where n is a constant and x is a variable; a decreasing geometric series has r: r<0).

I take back my statement that it was an exponential function.

x^n is not an exponential function, n^x is.

I can prove the convergency of an exponetial function, but I did not change sin(1/x) into and exponential. The form I turned the equation into was a monomial with degree of n, because this is easier to determin the convergency of.

Sorry for the mix up.

Originally posted by econundrum
I take back my statement that it was an exponential function.

x^n is not an exponential function, n^x is.

I can prove the convergency of an exponetial function, but I did not change sin(1/x) into and exponential. The form I turned the equation into was a monomial with degree of n, because this is easier to determin the convergency of.

Sorry for the mix up.

Maybe this can help you:

sin x = x - x^3 / 3! + x^5 / 5! - x^7 / 7! + ....

The problem in divergence is not in te even terms of the sum, as sin(1/n) &lt; 1/n for n&gt;1.

So sin(1/2) + sin (1/4) + ... &lt; 1/2 + 1/4 + ... = 1

What's left is the sum for odd n of sin(1/n)

there are some assertions you make which are incorrect, and that may be what is throwing you off.
frist of all, infinity^0 is not 1, it is &quot;undefined&quot; (can be anything).
second, lim x-&gt;0 sin x = x so, as n gets large, sin (1/n) approaches 1/n--and as we all know, sum(n=1 to inf) 1/n is unbounded (it is called the &quot;harmonic series&quot; and is useful for a number of applications but does not converge).
therefore, since the terms actually are a little larger than 1/n, the sum cannot converge.

another mistake

lim (x-&gt;infinity) ln(sin(1/x))/ln(1/x) = -infinity/infinity

Originally posted by BarefootChessPlayer
frist of all, infinity^0 is not 1, it is "undefined" (can be anything).
second, lim x->0 sin x = x so, as n gets large, sin (1/n) approaches 1/n--and as we all know, sum(n=1 to inf) 1/n is unbounded (it is called the "harmonic series" and is useful for a number of applications but does not converge).
therefore, since the terms actually are a little larger than 1/n, the sum cannot converge.

First of all, thank you for the first fact you pointed out, that infinity^0 is not 1. I had heard that before, but I had never seen a situation where infinity^0 was not 1, so I had assumed it to be wrong. (My teacher didn't even catch that).

It is the second thing you said though that stumps me.

How does the lim x-&gt;0 sin x equal to x. I have never heard that before. If it is true (I don't doubt you, I just don't know why it is true) than it definately helps in my proof (and I do know what a harmonic series is, but thanks anyways).

P.S. thanks to everyone who replied with useful information! It definately helped a lot.

Look at my first post. For x -&gt;0 the terms with x^3, x^5, x^7, ... become neglectable in comparison to the first term, x. So for x-&gt;0 sin(x)=x

With taylor:
sin(x) &gt;= sin(0) + sin'(0)*x + sin''(0)*(x^2)/2 + (-1)*(x^3)/6
sin(x) &gt;= x-(x^3)/6

and also:
sin(x) &lt;= sin(0) + 1*x
sin(x) &lt;= x

So x-(x^3)/6 &lt;= sin(x) &lt;= x

for x near 0 x^3 becomes insignificant to x.

more generaly:
sum{k=0 to n}((-1)^n*(x^(2k+1))/(2k+1)!) &lt;= sin(x) if n even
sum{k=0 to n}((-1)^n*(x^(2k+1))/(2k+1)!) &gt;= sin(x) if n odd

note that sum{k=0 to n}((-1)^n*(x^(2k+1))/(2k+1)!)=sin(x). It difficult to see here in this horrible (formal) notation.

Originally posted by Fiathahel
With taylor:
sin(x) >= sin(0) + sin'(0)*x + sin''(0)*(x^2)/2 + (-1)*(x^3)/6
sin(x) >= x-(x^3)/6

and also:
sin(x) <= sin(0) + 1*x
sin(x) <= x

So x-(x^3)/6 <= sin(x) <= x

for x near 0 x^3 becomes insignificant to x.

more generaly:
sum{k=0 to n}((-1)^n*(x^(2k+1))/(2k+1)!) <= sin(x) if n even
sum{k=0 to n}((-1)^n*(x^(2k+1))/(2k+1)!) >= sin ...[text shortened]... ((-1)^n*(x^(2k+1))/(2k+1)!)=sin(x). It difficult to see here in this horrible (formal) notation.

Lol, that's why i wrote it out with dots in my post 🙂

Originally posted by econundrum
First of all, thank you for the first fact you pointed out, that infinity^0 is not 1. I had heard that before, but I had never seen a situation where infinity^0 was not 1, so I had assumed it to be wrong. (My teacher didn't even catch ...[text shortened]... who replied with useful information! It definately helped a lot.

the second part has been addressed in a previous post so i'll not go into that now.
you might think of &quot;x raised to the zero power&quot; as &quot;x/x&quot;. it's unity for all x except x = 0 and x = inf, where it is undefined because the quotient can be anything. (again, &quot;zero times what equals zero?&quot; and &quot;infinity times what equals infinity?&quot; ).
here are three examples to show why inf/inf cannot be assigned a specific value.
1. lim x-&gt;+inf (3^x)/(2^x) = + infinity becaue it simplifies to (3/2)^x.
2. the reciprocal of the above, lim x-&gt;0 2^x/3^x = 0, as (2/3)^inf is zero.
3.lim x-&gt;+inf exp(x + 2)/exp(x) = exp(2) (e^2).
since we have found three different values for inf/inf, that quantity cannot be assigned a definite value.
for inf^0, one example that comes to mind is (exp(x)^(1/x)), which goes to exp(1).
hope this isn't too confusing.

Thank you guys. At first I wasn't able to understand completely some of your posts, and so I had to take a second look at some of them. Now that I do understand them, I realize you guys were right, and I also realize how easy it was to solve the equation (much easier than I had made it).
I'm glad I made my orriginal post, and, again, thanks for all of your help.
Econundrum