Series 215 problem 01.mws

1 . Series and partial sums.

A series is a sequence of partial sums, hence a series converges if and only if its sequence of partial sums converges. We explore this more here. Consider the geometric series Sum(1/exp(2*n),n = 1 .. infinity) . First the terms of the sequence to be added.

> a := n-> 1/exp(2*n);
Now we build the sequence of partial sums associated to the series in Exercise 14.

a := proc (n) options operator, arrow; 1/exp(2*n) e...

> S := n-> sum(a(k),k=1..n);

And for example we can obtain the first 30 terms of both sequences:

S := proc (n) options operator, arrow; sum(a(k),k =...

> seq(a(n),n=1..20);

1/exp(2), 1/exp(4), 1/exp(6), 1/exp(8), 1/exp(10), ...

> seq(evalf(S(n)),n=1..30);

.1353352832, .1536509221, .1561296743, .1564651369,...
.1353352832, .1536509221, .1561296743, .1564651369,...
.1353352832, .1536509221, .1561296743, .1564651369,...

You should be sure to understand where both these sequences come from. From this work, it looks like the series converges to .1565176426 (at least to 10 significant digits). In fact, we see that this series is a geometric series with a = 1/exp(2) and r = 1/exp(2) . Since abs(r) < 1 , we have by the geometric series formula, that this series will sum to exp(-2)/(1-exp(-2)) = 1/(exp(2)-1) which is to 10 significant digits equal to .1565176427.

> S(n);

-exp(1)^2/(-1+exp(1)^2)/exp(n+1)^2+1/(-1+exp(1)^2)

> Limit(Sum(a(k),k=1..n),n=infinity) = limit((S(n),n=infinity));

Limit(Sum(1/exp(2*k),k = 1 .. n),n = infinity) = 1/...

Now Maple can do infinite series - but you should know that an infinite series is always a limit of partial sums as discussed here.

> limit(S(n),n=infinity);

1/(-1+exp(2))

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Submission:

Consider the series Sum(1/((3*n-2)*(3*n+1)),n = 1 .. infinity) .

(a) Argue from the test for divergence , why this series is POSSIBLY convergent.

(b) Use Maple's partial fraction command to rewrite the terms of the series.

(c) Use your result in (b) to find an expression for the n th term in the sequence of partial sums.

(d) Take the limit of this expression as proc (n) options operator, arrow; infinity end proc... , and confirm your answer with Maple.

(e) Consider the series Sum(ln(n/(n+1)),n = 1 .. infinity) . Determine whether this series converges. Justify your answer.

Submission worksheet:

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2. Plotting the n 'th partial sum of a series.

If s[n] = Sum(a[k],k = 1 .. n) , then we can plot the sequence of partial sums in order to get an impression of whether the series converges. For example, consider the series Sum(1/sqrt(k)-1/sqrt(k+1),k = 1 .. infinity) . This time, let us define the k 'th term and n 'th partial sum as functions, and then plot the first few partial sums.

> a:=k->1/sqrt(k)-1/sqrt(k+1);

a := proc (k) options operator, arrow; 1/sqrt(k)-1/...

> s:=n->sum(a(k),k=1..n);

s := proc (n) options operator, arrow; sum(a(k),k =...

> plot([seq([n,s(n)],n=1..1000)],style=point);

[Maple Plot]

The plot does support the idea that the sequence of partial sums may be converging to 1. But pictures can be very deceiving, so be careful. For example, consider the series represented below.

> a:=k->1/k;

a := proc (k) options operator, arrow; 1/k end proc...

> s:=n->sum(a(k),k=1..n);

s := proc (n) options operator, arrow; sum(a(k),k =...

> plot([seq([n,s(n)],n=1..1000)],style=point);

[Maple Plot]

The sum of the first 1000 terms is not very large. But we know the series diverges, because this is just the harmonic series. Even Maple knows this series diverges.

> sum(a(k),k=1..infinity);

infinity

Submission:

Consider the series given by a[k] = 1/(k^3*sin(k)^2) .

a) Have Maple try to compute Limit(s(n),n = infinity) . If that does not work, ask Maple for a floating point estimate of the sum. What does Maple's response tell you?

b) Plot the first 100 points in the sequence of partial sums. Use this to give an estimate of the sum of the series.

c) Plot the first 200 points in the sequence of partial sums. Use this to give a new estimate of the sum of the series.

d) Plot the first 400 points in the sequence of partial sums. What do you observe now?

e) You probably have observed that there is some interesting behaviour in the sequence of partial sums that happens when n = 355 . Find a floating point estimate of the number 355/113, and explain why that estimate has something to do with what happens.

Submission Worksheet:

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