# How do you evaluate the sum represented by #sum_(n=1)^5n/(2n+1)# ?

Next, perform the operations in the denominator...

Next, we have to multiply each numerator and denominator by the missing components...

#2/5*693/693=1386/3465,
3/7*495/495=1485/3465,
4/9*385/385=1540/3465
and 5/11*315/315=1575/3465#

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To evaluate the sum ( \sum_{n=1}^{5} \frac{n}{2n+1} ), follow these steps:

- Substitute ( n ) with each value from 1 to 5.
- Calculate the corresponding fraction for each value of ( n ).
- Add up all the fractions.

Here's the breakdown:

For ( n = 1 ): ( \frac{1}{2(1) + 1} = \frac{1}{3} )

For ( n = 2 ): ( \frac{2}{2(2) + 1} = \frac{2}{5} )

For ( n = 3 ): ( \frac{3}{2(3) + 1} = \frac{3}{7} )

For ( n = 4 ): ( \frac{4}{2(4) + 1} = \frac{4}{9} )

For ( n = 5 ): ( \frac{5}{2(5) + 1} = \frac{5}{11} )

Now, add up all the fractions:

( \frac{1}{3} + \frac{2}{5} + \frac{3}{7} + \frac{4}{9} + \frac{5}{11} )

Find a common denominator for all fractions, which is 3465.

( = \frac{1215}{3465} + \frac{1386}{3465} + \frac{1470}{3465} + \frac{1556}{3465} + \frac{1505}{3465} )

Add the numerators:

( = \frac{1215 + 1386 + 1470 + 1556 + 1505}{3465} )

( = \frac{7132}{3465} )

Therefore, ( \sum_{n=1}^{5} \frac{n}{2n+1} = \frac{7132}{3465} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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