Sequences and Series: 1/[(k+2)(k+3)], 1/(k-2)

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warwick

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I heard these would give me trouble. For various reasons, I have been extremely scatterbrained lately. My notes are incomplete and of no help. I honestly don't know where to begin.

Determine whether the series converges, and if so find its sum.

7) Sigma, k=1 to infinity 1/[(k+2)(k+3)]

9) Sigma, k=3 to infinity 1/(k-2)
 
Re: Sequences and Series

warwick said:
I heard these would give me trouble. For various reasons, I have been extremely scatterbrained lately. My notes are incomplete and of no help. I honestly don't know where to begin.

Determine whether the series converges, and if so find its sum.

7) Sigma, k=1 to infinity 1/[(k+2)(k+3)]

This is called a telescoping series. If expand it you find

1/[(k+2)(k+3)] = 1/(k+2) - 1/(k+3)

so:

Sigma, k=1 to infinity 1/[(k+2)(k+3)]

= (1/3 - 1/4) + (1/4 - 1/5) + (1/5 - 1/6) + (1/6 - 1/7) + ..... see the pattern?

9) 7) Sigma, k=3 to infinity 1/(k-2)

Start writing the first few terms - familiar series should pop-up.

This is called a "harmonic" series. If you don't know about this type of series, do a google search and tell us what you found
Click to expand...
 
Ok. The book says harmonic series diverge. Do I just write the series diverges by observation of the type of series?
 
Use a-sub n+1 / a-sub to show that the given sequence {a-sub n} is strictly increasing or strictly decreasing.

9) {ne^(-n)} from n=1 to +infinity

11) {(n^n)/n!} from n=1 to +infinity
 
Show some work - so that we know where to begin.
 
Subhotosh Khan said:
Show some work - so that we know where to begin.
Click to expand...

My first question is are some of these concluded by observation? For #9, I eventually got

(n+1) e^n / n e^(n+1)

and by observation realized that the denominator will always be bigger than the numerator.
 
\(\displaystyle \L 9)\;\frac{{\frac{{n + 1}}{{e^{n + 1} }}}}{{\frac{n}{{e^n }}}} = \frac{{n + 1}}{n}\frac{{e^n }}{{e^{n + 1} }} = \left( {1 + \frac{1}{n}} \right)\left( {\frac{1}{e}} \right)\)
 
pka said:
\(\displaystyle \L 9)\;\frac{{\frac{{n + 1}}{{e^{n + 1} }}}}{{\frac{n}{{e^n }}}} = \frac{{n + 1}}{n}\frac{{e^n }}{{e^{n + 1} }} = \left( {1 + \frac{1}{n}} \right)\left( {\frac{1}{e}} \right)\)
Click to expand...

Ok. You broke up some fractions and did some trickery, as my professor says. Haha. Got it.
 
Sigma, k=3 to infinity 1/(k-2)

Do I just write this diverges by the fact of it being a harmonic series or can I work it out?
 
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