@Satvik Golechha
–
Look up "Herschfeld's Convergence Theorem". It states that for real terms \({ { x }_{ n } }\ge 0\) and real powers \(1>p>0\) if and only if

\(\displaystyle{ { x }_{ n } }^{ { p }^{ n } }\)

is bounded, then the following converges

\({ x }_{ 0 }+{ \left( { x }_{ 1 }+{ \left( { x }_{ 2 }+{ \left( ...+{ \left( { x }_{ n } \right) }^{ p } \right) }^{ p } \right) }^{ p } \right) }^{ p }\)

as \(n\rightarrow \infty \)

Here, \(p=\frac { 1 }{ 2 } \), so it's easy to see how the condition is met.
–
Michael Mendrin
·
2 years, 2 months ago

Log in to reply

Comment deleted
Aug 28, 2015

Log in to reply

@Naman Kapoor
–
Is it a typo where \(r\) should be \(i\)? But even so, the terms approach \(1\) for large \(i\), so this doesn't converge.
–
Michael Mendrin
·
1 year, 6 months ago

## Comments

Sort by:

TopNewestThis is the "nested radical constant", also known as Vijayaraghavan's constant, which is \(1.75793275...\)

There is no exact expression for this constant. – Michael Mendrin · 2 years, 2 months ago

Log in to reply

– Satvik Golechha · 2 years, 2 months ago

Sir, how do we find out whether a sum like this converges or not? Thanks.Log in to reply

\(\displaystyle{ { x }_{ n } }^{ { p }^{ n } }\)

is bounded, then the following converges

\({ x }_{ 0 }+{ \left( { x }_{ 1 }+{ \left( { x }_{ 2 }+{ \left( ...+{ \left( { x }_{ n } \right) }^{ p } \right) }^{ p } \right) }^{ p } \right) }^{ p }\)

as \(n\rightarrow \infty \)

Here, \(p=\frac { 1 }{ 2 } \), so it's easy to see how the condition is met. – Michael Mendrin · 2 years, 2 months ago

Log in to reply

Log in to reply

– Michael Mendrin · 1 year, 6 months ago

Is it a typo where \(r\) should be \(i\)? But even so, the terms approach \(1\) for large \(i\), so this doesn't converge.Log in to reply