If m = 1+(1/root3)+(1/root 5)+(1/root7)+........... and
n = 1+(1/root2)+(1/root 4)+(1/root6)+...
Then find m+n and m-n??
If m = 1+(1/root3)+(1/root 5)+(1/root7)+........... and
n = (1/root2)+(1/root 4)+(1/root6)+...
Then find m+n and m-n??
why post it twice?
if these series go on forever, they both diverge, since they are both greater than the Harmonic Series.
To find the values of m and n, we need to understand the patterns they follow.
For m, we have a series of terms where each term is the reciprocal of the square root of an odd number. So, the first term is 1/root3, the second term is 1/root5, and so on. We can see that the denominator of each term is increasing by 2. Therefore, the pattern is that the nth term is given by 1/root(2n+1).
To find the value of m, we need to sum all these terms. Let's start by representing the sum of m as S_m:
S_m = 1/root3 + 1/root5 + 1/root7 + ...
Now, let's multiply each term in S_m by its conjugate to rationalize the denominators. For example:
1/root3 * root3/root3 = root3/3
Doing this for each term, we get:
S_m = (1/root3) * (root3/ root3) + (1/root5) * (root5/ root5) + (1/root7) * (root7/ root7) + ...
= root3/3 + root5/5 + root7/7 + ...
Notice that we can factor out the square root from each term:
S_m = (root3/3) + (root5/5) + (root7/7) + ...
= root3/3 + root5/5 + root7/7 + ...
To find the value of n, we have a similar pattern, except the denominators are increasing by 2 starting from 2. Therefore, the nth term is given by 1/root(2n).
Following the same steps as above, we can represent the sum of n as S_n:
S_n = 1/root2 + 1/root4 + 1/root6 + ...
By rationalizing the denominators, we get:
S_n = (1/root2) * (root2/root2) + (1/root4) * (root4/root4) + (1/root6) * (root6/root6) + ...
= root2/2 + root4/4 + root6/6 + ...
And again, factoring out the square root from each term:
S_n = (root2/2) + (root4/4) + (root6/6) + ...
= root2/2 + root4/4 + root6/6 + ...
Now, to find the values of m and n, we cannot determine an exact numeric value, but we can simplify them as follows:
m = root3/3 + root5/5 + root7/7 + ...
n = root2/2 + root4/4 + root6/6 + ...
To find m + n, we add these two expressions:
m + n = (root3/3 + root5/5 + root7/7 + ...) + (root2/2 + root4/4 + root6/6 + ...)
= root3/3 + root5/5 + root7/7 + ... + root2/2 + root4/4 + root6/6 + ...
To find m - n, we subtract the expressions:
m - n = (root3/3 + root5/5 + root7/7 + ...) - (root2/2 + root4/4 + root6/6 + ...)
= root3/3 + root5/5 + root7/7 + ... - (root2/2 + root4/4 + root6/6 + ...)