Euler And Series

Yahoo Answers is a goldmine. There are a lot of great questions in there. Today I have one relating to series and Euler. Yes, this guy again. He always comes back for more.

A sequence a_n is defined as 1/n²(n + 1)². Find the sum of the series from 1 to infinity.

Remember Euler from before? He proved that the sum of 1/n² from 1 to infinity is π²/6. Yes, Euler was a smart man. All we need now is to transofrm a_n to something with 1/n² which includes sums. But how?

Let's start with the obvious and proceed from there. To create that denominator, we can try to add 1/n² and 1/(n + 1)², see what we get, and then decide what else to do. So:
1/n² + 1/(n + 1)² = (n + 1)²/n²(n + 1)² + n²/n²(n + 1)²

Which is:
(n² + (n + 1)²)/n²(n + 1)²

Open parentheses and add:
(n² + n² + 2n + 1)/n²(n + 1)²
(2n² + 2n + 1)/n²(n + 1)²

To leave only the 1 in the numerator (which will give a_n, we need to subtract (2n² + 2n)/n²(n + 1)²:
a_n = (2n² + 2n + 1)/n²(n + 1)² - (2n² + 2n)/n²(n + 1)²

The second term can be also written as 2n(n + 1), which can be cancelled. At the end, we get:
a_n = (2n² + 2n + 1)/n²(n + 1)² - 2/n(n + 1)

2/n(n + 1) can also be written as (2n + 2 - 2n)/n(n + 1), which can be transformed into:
(2n + 2)/n(n + 1) - 2n/n(n + 1) = 2/n - 2/(n + 1)

Since it's the opposite on a_n, we can finally conclude that:
a_n = 1/n² + 1/(n + 1)² + 2/(n + 1) - 2/n

Now it's really easy. Notice how 2/(n + 1) and -2/n cancel each other, except for the first term. Their sum is -2/1 + 2/2 - 2/2 + 2/3 - 2/3 + 2/4 - 2/4 + ... . Since 2/∞ is zero, the only remainder of this part of the series is -2.

As Euler said, the sum of the 1/n² part of the series is π²/6. The 1/(n + 1)² has the exact same sum, but minus one. Since the sums starts from 1/2² and not 1/1², the 1/1² is left outside.

The total sum of a_n from 1 to infinity is:
π²/6 + π²/6 - 1 - 2
S = π²/3 - 3

Since the terms of a_n get very small very quickly, it can be easily approximated for validity.

Hope you like it,
Nadav

nadavs

Limiting Euler

Today's question is about limits and Euler's number, e. This number has many interesting properties, and one of them is the derivative of e^x, which is e^x. Today's question does not deal with the derivative of this function, but with some limits.

Find the limit:
lim_{x -> infinity} (e^x + x)^1/x

This question is definitely not easy, mainly because of that annoying x in there and the 1/x exponent. However, we must never give up. We need to fight this question, so let's fight it with simplicity. Define another variable!

y = (e^x + x)^1/x

Simple, isn't it? All we have to do now is... well... find the limit of y. That didn't help. You can see there is a fraction in the exponent, so if we could just bring it down to a normal fraction, we could use L'Hopital's rule and solve this.

To get the exponent down, we can use a very well known technique: logs. There is a great rule in logarithms which states:
log_a bⁿ = nlog_a b

Using that fantastic rule we can bring the 1/x down and use L'Hopital's rule! To get a nice answer, let's take the logarithm of base e of this function, also written as ln (natural logarithm). So:
ln y = ln (e^x + x)^1/x

Using the logarithm rule mentioned earlier, we can say this equals:
ln (e^x + x) * 1/x
ln (e^x + x) / x

Mr. L'Hopital starts to smile. Now we can find the limit of this function and remember this is the logarithm of the function, so we'll need to consider that.

lim_{x -> infinity} ln y = lim_{x -> infinity} ln (e^x + x) / x

Using L'Hopital's rule (lim_{x -> a} f(x)/g(x) = lim_{x -> a} f'(x)/g'(x)), we can conclude that:
lim_{x -> infinity} ln (e^x + x) / x
is equal to:
lim_{x -> infinity} (e^x + 1) * (1 / (e^x + x)) / 1
lim_{x -> infinity} (e^x + 1) / (e^x + x)

Use L'Hopital's rule twice more:
lim_{x -> infinity} e^x / (e^x + 1)
lim_{x -> infinity} e^x / e^x

Now that looks familiar, doesn't it? This limit equals 1.

Remember: lim_{x -> infinity} ln y = 1, so lim_{x -> infinity} y = e

The limit of the entire function is e. Problem solved.

Hope you enjoyed,
Nadav

nadavs