# What is limit (1-1/3+1/5-1/7+...+(-1)^(n-1)1/(2n-1))? n go infanswer is pie /4

*print*Print*list*Cite

You need to use squeeze theorem to evaluate the given limit, hence, considering the following function `f_(n) (x) = x^n/(x^2+1)` , you may evaluate `I_0` such that:

`I_0 = int_0^1 1/(x^2 + 1) dx = (arctanx)|_0^1`

Using the fundamental theorem of calculus yields:

`I_0 = arctan 1 - arctan 0 => I_0 = pi/4 - 0 => I_0 = pi/4`

You need to evaluate `I_1` such that:

`I_1 = int_0^1 x/(x^2 + 1) dx`

You need to use the substitution process to evaluate the definite integral `I_1` such that:

`x^2+1 = t => 2xdx = dt => xdx = (dt)/2`

Changing the variable yields:

`int_1^2 ((dt)/2)/t =(1/2) ln|t||_1^2 = (1/2) ln 2 `

You need to evaluate `I_2` such that:

`I_2 = int_0^1 x^2/(x^2 + 1) dx`

Adding and subtracting 1 yields:

`I_2 = int_0^1 (x^2+1-1)/(x^2+1) dx`

Using the linearity of integral yields:

`I_2 = int_0^1 (x^2+1)/(x^2+1) dx - int_0^1 1/(x^2+1) dx `

`I_2 = x|_0^1 - I_0`

`I_2 = 1 - I_0`

`I_(2*1) = 1/(2*1-1) - I_(2*1-2)`

You may state that the reccurence relation is the following, such that:

`I_(2n) = 1/(2n-1) - I_(2n-2) => 1/(2n-1) = I_(2n) + I_(2n-2)`

You may write the given summation using the reccurence relation such that:

`1 - 1/3 + 1/5 - 1/7 + ... + (-1)^(n-1)*1/(2n-1) = I_0 + I_2 - (I_2 + I_4) + (I_4 + I_6) - (I_6 + I_8) + ... + (-1)^(n-1)(I_(2n-2) + I_(2n))`

Reducing duplicate terms yields:

`1 - 1/3 + 1/5 - 1/7 + ... + (-1)^(n-1)*1/(2n-1) = I_0 + (-1)^(n-1)I_(2n)`

You should notice that the following aspect of the function `f_(2n)(x) = (x^(2n))/(x^2+1)` such that:

`0 <= (x^(2n))/(x^2+1) <= x^(2n) if x in [0,1]`

Integrating the inequality, yields:

`0 <= int_0^1 (x^(2n))/(x^2+1) dx <= int_0^1 x^(2n) dx`

`0 <= I_(2n) <= (x^(2n+1))/(2n+1)|_0^1`

`0 <= I_(2n) <= 1/(2n+1)`

Evaluating the limit yields:

`lim_(n->oo) 0 <= lim_(n->oo) I_(2n) <= lim_(n->oo) 1/(2n+1)`

`0 <= lim_(n->oo) I_(2n) <= 0`

Using the squeeze theorem yields that `lim_(n->oo) I_(2n) = 0` , hence, since `|(-1)^(n-1)I_(2n)| = I_(2n) => lim_(n->oo)(-1)^(n-1)I_(2n) = 0` .

You may evaluate the limit of the given summation such that:

`lim_(n->oo)(1 - 1/3 + 1/5 - 1/7 + ... + (-1)^(n-1)*1/(2n-1)) = lim_(n->oo)(I_0 + (-1)^(n-1)I_(2n))`

`lim_(n->oo)(1 - 1/3 + 1/5 - 1/7 + ... + (-1)^(n-1)*1/(2n-1)) = lim_(n->oo)I_0 + lim_(n->oo) (-1)^(n-1)I_(2n)`

`lim_(n->oo)(1 - 1/3 + 1/5 - 1/7 + ... + (-1)^(n-1)*1/(2n-1)) = I_0 + 0`

`lim_(n->oo)(1 - 1/3 + 1/5 - 1/7 + ... + (-1)^(n-1)*1/(2n-1)) = pi/4`

**Hence, evaluating the given limit, using the function `f_(n) = x^n/(x^2+1)` and the squeeze theorem, yields that `lim_(n->oo)(1 - 1/3 + 1/5 - 1/7 + ... + (-1)^(n-1)*1/(2n-1)) = pi/4` .**

lim_(n->oo)