How do you use the summation formulas to rewrite the expression #Sigma (2i+1)/n^2# as i=1 to n without the summation notation and then use the result to find the sum for n=10, 100, 1000, and 10000?

Question

Paisley Johnson · Accepted Answer

# sum_(i=1)^n (2i+1)/n^2 = (n+2)/n #

Let # S_n = sum_(i=1)^n (2i+1)/n^2 #
# :. S_n = 1/n^2sum_(i=1)^n (2i+1) #
# :. S_n = 1/n^2{2sum_(i=1)^n (i)+sum_(i=1)^n (1)} #

And using the standard results:
# sum_(r=1)^n r = 1/2n(n+1) #

We have;

# \ \ \ \ \ S_n = 1/n^2{2*1/2n(n+1) + n} #
# :. S_n = 1/n^2{n(n+1) + n} #
# :. S_n = 1/n^2{n^2+n + n} #
# :. S_n = 1/n^2{n^2+2n} #
# :. S_n = 1/n^2{n(n+2)} #
# :. S_n = (n+2)/n #

And this has been calculated using Excel for #n=10, 100, 1000, 10000#

What happens as #n rarr oo#?

[ NB As an additional task we could possibly conclude that as #n rarr oo# then #S_n rarr 1#; This is probably the conclusion of this question]

Now, # S_n = (n+2)/n #

# :. S_n = 1+2/n #

And so,

# lim_(n rarr oo) S_n = lim_(n rarr oo) (1+2/n) #
# :. lim_(n rarr oo) S_n = lim_(n rarr oo) (1) + 2lim_(n rarr oo)(1/n) #
# :. lim_(n rarr oo) S_n = 1+0 #
# :. lim_(n rarr oo) S_n = 1#
Which confirms our assumption!

Scarlett Allison · Answer

undefined To rewrite the expression $ \sum_{i=1}^{n} \frac{2i+1}{n^2} $ without summation notation, you can use the property of linearity of summation. Then, you can find the sum by plugging in the values of $ n $ as 10, 100, 1000, and 10000 into the resulting expression.

The expression can be rewritten as:

$$ \frac{2}{n^2} \sum_{i=1}^{n} i + \frac{1}{n^2} \sum_{i=1}^{n} 1 $$

$$ = \frac{2}{n^2} \left( \frac{n(n+1)}{2} \right) + \frac{1}{n^2} \left( \sum_{i=1}^{n} 1 \right) $$

$$ = \frac{n+1}{n} + \frac{1}{n^2} \left( n \right) $$

$$ = \frac{n+1}{n} + \frac{1}{n} $$

$$ = 1 + \frac{1}{n} $$

Using this expression, we can find the sum for $ n = 10, 100, 1000, $ and $ 10000 $:

For $ n = 10 $:
$$ = 1 + \frac{1}{10} = 1.1 $$

For $ n = 100 $:
$$ = 1 + \frac{1}{100} = 1.01 $$

For $ n = 1000 $:
$$ = 1 + \frac{1}{1000} = 1.001 $$

For $ n = 10000 $:
$$ = 1 + \frac{1}{10000} = 1.0001 $$