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# 6.7 Calculation of

6.81   Example. We will calculate , where is a positive number. Let . Then (see the figure)

and hence

Thus

i.e.
 (6.82)

Since

it follows from the squeezing rule that
 (6.83)

Notice that in this example the squeezing rule has allowed us to prove the existence of a limit whose existence was not obvious.

6.84   Example. We will show that for all
 (6.85)

Let , and let . Let

By (6.82), we have

so

It follows from (6.83) that
 (6.86)

From the picture, we see that

i.e.

Hence

By (6.86), we have

so by the squeezing rule, , i.e.

This completes the proof of (6.85).

The reason we assumed to be positive in the previous example was to guarantee that has a logarithm. We could extend this proof to work for arbitrary , but we suggest an alternate proof for negative in exercise 6.97

6.87   Example (Numerical calculation of ) It follows from the last example that

I wrote a Maple procedure to calculate by using this fact. The procedure limcalc(n) below calculates

and I have printed out the results for n = 1,2,...,6.

> limcalc := n -> (1+ .01^n)^(100^n);

> limcalc(1);

> limcalc(2);

> limcalc(3);

> limcalc(4);

> limcalc(5);

> limcalc(6);

6.88   Exercise. From my computer calculations it appears that

Explain what has gone wrong. What can I conclude about the value of from my program?

6.89   Example. Actually, Maple is smart enough to find the limit, and does so with the commands below. The command evalf returns the decimal approximation of its argument.

> limit( (1+1/n)^n,n=infinity);

> evalf(%);

6.90   Entertainment ( .) Find the limit of the sequence , or else show that the sequence diverges.

6.91   Example (Compound interest.) The previous exercise has the following interpretation.

Suppose that dollars is invested at % annual interest, compounded times a year. The value of the investment at any time is calculated as follows:

Let , and let be the value of the investment at time Then

 (6.92)

and in general
 (6.93)

The value of the investment does not change during the time interval
For example, if denotes the value of one dollar invested for one year at % annual rate of interest with the interest compounded times a year, then

Thus it follows from our calculation that if one dollar is invested for one year at % annual rate of interest, with the interest compounded infinitely often'' or continuously'', then the value of the investment at the end of the year will be

If the rate of interest is 100%, then the value of the investment is dollars, and the investor should expect to get \$2.71 from the bank.

This example was considered by Jacob Bernoulli in 1685. Bernoulli was able to show that .[8, pp94-97]

6.94   Exercise. Calculate the following limits.
A
a)
.
b)
.

6.95   Exercise.
a)
Use the formula for a finite geometric series,

to show that
 (6.96)

b)
Let Use inequality (6.96) to show that

for all such that .
c)
Prove that for all .

6.97   Exercise. A Let . Use exercise 6.95 to show that

(Hence we have .)

Hint: Note that for all real numbers .

Next: 7. Still More Area Up: 6. Limits of Sequences Previous: 6.6 Geometric Series   Index
Ray Mayer 2007-09-07