Question :
sum _n=1^infinity frac(-1)^nln n+1
Determine the convergence or divergence
Converges
Evaluate
n=1∑+∞ln(n)+1(−1)n
Find the limit
n→+∞lim(ln(n)+1(−1)n)
Remove the absolute value bars
n→+∞lim(ln(n)+11)
Rewrite the expression
limn→+∞(ln(n)+1)1
Calculate
More Steps
Evaluate
n→+∞lim(ln(n)+1)
Rewrite the expression
n→+∞lim(ln(n))+n→+∞lim(1)
Calculate
More Steps
Evaluate
n→+∞lim(ln(n))
Rewrite the expression
ln(n→+∞lim(n))
Calculate
ln(+∞)
Calculate
+∞
(+∞)+n→+∞lim(1)
Calculate
(+∞)+1
Calculate
+∞
+∞1
Calculate
0
Rewrite the expression
−ln(n+1)+1ln(n)+1>−1
Change the signs on both sides of the inequality and flip the inequality sign
ln(n+1)+1ln(n)+1<1
Calculate
ln(n+1)+1ln(n)+1−1<0
Calculate
More Steps
Calculate
ln(n+1)+1ln(n)+1−1
Reduce fractions to a common denominator
ln(n+1)+1ln(n)+1−ln(n+1)+1ln(n+1)+1
Write all numerators above the common denominator
ln(n+1)+1ln(n)+1−(ln(n+1)+1)
Calculate the sum or difference
More Steps
Evaluate
ln(n)+1−(ln(n+1)+1)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
ln(n)+1−ln(n+1)−1
Use logax−logay=logayx to transform the expression
ln(n+1n)+1−1
Since two opposites add up to 0,remove them form the expression
ln(n+1n)
ln(n+1)+1ln(n+1n)
ln(n+1)+1ln(n+1n)<0
Separate the inequality into 2 possible cases
{ln(n+1n)>0ln(n+1)+1<0{ln(n+1n)<0ln(n+1)+1>0
Solve the inequality
More Steps
Evaluate
ln(n+1n)>0
For e>1 the expression ln(n+1n)>0 is equivalent to n+1n>e0
n+1n>e0
Evaluate the power
n+1n>1
Calculate
n+1n−1>0
Calculate
More Steps
Calculate
n+1n−1
Reduce fractions to a common denominator
n+1n−n+1n+1
Write all numerators above the common denominator
n+1n−(n+1)
Subtract the terms
n+1−1
Use b−a=−ba=−ba to rewrite the fraction
−n+11
−n+11>0
Change the signs on both sides of the inequality and flip the inequality sign
n+11<0
Rewrite the expression
n+1<0
Move the constant to the right side
n<0−1
Removing 0 doesn't change the value,so remove it from the expression
n<−1
{n<−1ln(n+1)+1<0{ln(n+1n)<0ln(n+1)+1>0
Solve the inequality
More Steps
Evaluate
ln(n+1)+1<0
Add or subtract both sides
ln(n+1)<0−1
Calculate
ln(n+1)<−1
For e>1 the expression ln(n+1)<−1 is equivalent to n+1<e−1
n+1<e−1
Evaluate the power
n+1<e1
Move the constant to the right side
n<e1−1
Subtract the numbers
More Steps
Evaluate
e1−1
Reduce fractions to a common denominator
e1−ee
Write all numerators above the common denominator
e1−e
n<e1−e
{n<−1n<e1−e{ln(n+1n)<0ln(n+1)+1>0
Solve the inequality
More Steps
Evaluate
ln(n+1n)<0
For e>1 the expression ln(n+1n)<0 is equivalent to n+1n<e0
n+1n<e0
Evaluate the power
n+1n<1
Calculate
n+1n−1<0
Calculate
More Steps
Calculate
n+1n−1
Reduce fractions to a common denominator
n+1n−n+1n+1
Write all numerators above the common denominator
n+1n−(n+1)
Subtract the terms
n+1−1
Use b−a=−ba=−ba to rewrite the fraction
−n+11
−n+11<0
Change the signs on both sides of the inequality and flip the inequality sign
n+11>0
Rewrite the expression
n+1>0
Move the constant to the right side
n>0−1
Removing 0 doesn't change the value,so remove it from the expression
n>−1
{n<−1n<e1−e{n>−1ln(n+1)+1>0
Solve the inequality
More Steps
Evaluate
ln(n+1)+1>0
Add or subtract both sides
ln(n+1)>0−1
Calculate
ln(n+1)>−1
For e>1 the expression ln(n+1)>−1 is equivalent to n+1>e−1
n+1>e−1
Evaluate the power
n+1>e1
Move the constant to the right side
n>e1−1
Subtract the numbers
More Steps
Evaluate
e1−1
Reduce fractions to a common denominator
e1−ee
Write all numerators above the common denominator
e1−e
n>e1−e
{n<−1n<e1−e{n>−1n>e1−e
Find the intersection
n<−1{n>−1n>e1−e
Find the intersection
n<−1n>e1−e
Find the union
n∈(−∞,−1)∪(e1−e,+∞)
Calculate
ln(n)+11
Calculate
ln(n+1)+11
The inequality is true
ln(n)+11>ln(n+1)+11
Solution
Converges
Show Solution
